The diagram below shows a representation of an electromagnetic wave.
Correctly label Figure 13 using the following symbols. (3 marks)
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Aussie Maths & Science Teachers: Save your time with SmarterEd
The diagram below shows a representation of an electromagnetic wave.
Correctly label Figure 13 using the following symbols. (3 marks)
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→ The wavelength is from the first peak on the electric wave to the second peak on the electric wave (not from an electric wave peak to magnetic wave peak).
Figure 8 shows a small ball of mass 1.8 kg travelling in a horizontal circular path at a constant speed while suspended from the ceiling by a 0.75 m long string.
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a.
b. \(1.2\ \text{ms}^{-1}\)
a. Only two forces are acting on the ball: \(F_w\) and \(F_T\):
b.
\(\text{Using}\ \ \tan\theta=\dfrac{F_{\text{net}}}{mg}:\)
\( F_{\text{net}}\) | \(=mg\,\tan\theta\) | |
\(\dfrac{mv^2}{r}\) | \(=mg\,\tan\theta\) | |
\(\dfrac{v^2}{r}\) | \(=g\,\tan\theta\) | |
\(\therefore v\) | \(=\sqrt{g\,r\,\tan\theta}\) | |
\(=\sqrt{9.8 \times 0.317 \times \tan 25}\ \ ,\ \ (r = 0.75 \times \sin25=0.317\ \text{m})\) | ||
\(=1.2\ \text{ms}^{-1}\) |
Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
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a. Decrease
b. Ammeter registers a current:
→ By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current.
→ This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
→ Hence by using the right-hand grip rule, the current will run clockwise through the coil.
c. → There will be an induced current from 6a to 6b.
→ As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
→ By the right-hand grip rule, the current will flow anti-clockwise around the coil.
→ There will also be an induced current from 6b to 6c.
→ The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
→ Using the right-hand grip rule, the current through the coil will flow clockwise.
a. Magnetic flux will decrease.
→ The magnetic flux is proportional to how many field lines are passing through a given area.
→ As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.
b. Ammeter registers a current:
→ By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current.
→ This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
→ Hence by using the right-hand grip rule, the current will run clockwise through the coil.
c. → There will be an induced current from 6a to 6b.
→ As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
→ By the right-hand grip rule, the current will flow anti-clockwise around the coil.
→ There will also be an induced current from 6b to 6c.
→ The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
→ Using the right-hand grip rule, the current through the coil will flow clockwise.
Consider the functions \(f\) and \(g\), where \begin{aligned} --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. \([-9, \infty)\) b. \(f\circ g(x)=x-9, \text{Domain}\ [0, \infty)\) c. \((-\infty, -3)\cap (3, \infty)\) a. \(\text{Range}\ \rightarrow\ [-9, \infty)\) \(g(x)=\sqrt{x} \ \rightarrow x\ \text{must be }\geq 0\) \(\therefore\ \text{Domain}\ f\circ g(x) \text{ is }[0, \infty)\) \(\text{For }g\circ h(x)\ \text{to exist}\ h(x)\geq 0\) \(x\text{-intercepts for }h(x)\ \text{are } x=-3, 3\) \(\text{and }h(x)\ \text{is positive for } x\leq -3\ \text{and }x\geq 3\) \(\therefore\ \text{Maximal domain} = (-\infty, -3)\cap (3, \infty)\)
& f: R \rightarrow R, f(x)=x^2-9 \\
& g:[0, \infty) \rightarrow R, g(x)=\sqrt{x}
\end{aligned}
b.
\(f\circ g(x)\)
\(=(g(x))^2-9\)
\(=(\sqrt{x})^2-9\)
\(=x-9\)
c.
\(g\circ h(x)\)
\(=\sqrt{h(x)}\)
\(=\sqrt{x^2-9}\)
The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit.
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a. \(r_{orbit}= 6.971 \times 10^6\ \text{m}\)
b. \(5780\ \text{s}\)
c. → The icon satellite is only subject to the gravitational force of the Earth.
→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit.
a. Radius of orbit is equal to the altitude plus the radius of the Earth.
\(r_{orbit}=600\ 000\ \text{m} + 6.371 \times 10^6\ \text{m} = 6.971 \times 10^6\ \text{m}\)
b. | \(\dfrac{r^3}{T^2}\) | \(=\dfrac{GM}{4\pi^2}\) |
\(T\) | \(=\sqrt{\dfrac{4\pi^2r^3}{GM}}\) | |
\(=\sqrt{\dfrac{4\pi^2 \times (6.971 \times 10^6)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}\) | ||
\(=5780\ \text{s}\ \ \text{(3 sig.fig.)}\) |
c. → The icon satellite is only subject to the gravitational force of the Earth.
→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit.
Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp. The cross-section of the ramp is modelled by the function \(f\), where \(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\) \(f(x)\) is both smooth and continuous at \(x=5\). The graph of \(y=f(x)\) is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres. --- 2 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below. --- 5 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- a. \(f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\) b.i \((25, 20)\) b.ii \([25, 55]\) b.iii \([5, 25]\) c. \(98.1\%\) d. \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\) \(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\) \(\text{making the trapezium rule approximation greater than the}\) \(\text{actual area.}\) e.i \(b=8.75, c=-283.4375\) e.ii \(34.10\ \text{cm}\) a. \(\text{Using CAS: Define f(x) then}\) \(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\) b.i \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\) \(\therefore\ \text{Point of inflection at }(25, 20)\) b.ii \(\text{Gradient function strictly increasing for }x\in [25, 55]\) b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\) c. \(\text{Using CAS:}\)
\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\) \(\text{making the trapezium rule approximation greater than the}\) \(\text{actual area.}\) e.i \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\) \(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\) \(\text{and }g'(55)=f'(55)\) \(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\) \(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\) \(\therefore b=\dfrac{35}{4}\quad (2)\) \(\text{Sub (2) into (1)}\) \(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\) \(c=-\dfrac{4535}{16}\) \(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\) e.ii \(\text{Using CAS: Solve }g(x)=0|x>55\) \(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)
\(\text{Area}\)
\(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
\(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)
\(\therefore\ \text{Estimate : Exact}\)
\(=637.5:650\)
\(=\dfrac{637.5}{650}\times 100\)
\(=98.0769\%\approx 98.1\%\)
d. \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)
\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\)
\(=\dfrac{35}{4}\)
\(c\)
\(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\)
\(=\dfrac{3165}{16}-55b\quad (1)\)
\(g'(x)=-\dfrac{x}{8}+b\)
Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).
Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
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a. Find \(v_0\) where forces due to magnetic and electric field are balanced
\(F_E\) | \(=F_B\) | |
\(qE\) | \(=qv_0B\) | |
\(v_0\) | \(=\dfrac{E}{B}\) |
b. \(v_0=2 \times 10^6\ \text{ms}^{-1}\)
c.i. Point \(\text{Z.}\)
c.ii. When electrons travel faster than \(v_0\):
→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged.
→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
a. Find \(v_0\) where forces due to magnetic and electric field are balanced
\(F_E\) | \(=F_B\) | |
\(qE\) | \(=qv_0B\) | |
\(v_0\) | \(=\dfrac{E}{B}\) |
b. \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
c.i. Point \(\text{Z.}\)
c.ii. When electrons travel faster than \(v_0\):
→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged.
→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
Quantised energy levels within atoms can best be explained by
\(D\)
→ This was explained by de Broglie’s work in determining electrons (all matter) can exhibit wave properties.
→ Quantised energy levels in the atom correspond to where electrons form standing waves due to there being an integer number of electron wavelengths.
\(\Rightarrow D\)
\(\text{I}\) | The energy of a light wave increases with increasing amplitude. |
\(\text{II}\) | The energy of a light wave increases with increasing frequency. |
\(\text{III}\) | The energy of a light wave increases with decreasing wavelength. |
\(D\)
→ Statement \(\text{I}\) is correct when considering the wave-model of light. The higher the amplitude, the further away the displacement of the wave from the origin which corresponds to higher energy.
→ Statements \(\text{II}\) and \(\text{III}\) are correct when considering the particle model of light.
→ The energy of a light photon, \(E=hf=\dfrac{hc}{\lambda}\). It follows that increasing the frequency or decreasing the wavelength both lead to an increase in energy.
\(\Rightarrow D\)
Matter is converted to energy by nuclear fusion in stars.
If the star Alpha Centauri converts mass to energy at the rate of 6.6 × 10\(^9\) kg s\(^{-1}\), then the power generated is closest to
\(C\)
→ The energy produced by the star each second is the power generated by the star.
→ \(E=mc^2=6.6 \times 10^9 \times (3 \times 10^8)^2=6 \times 10^{26}\ \text{Js}^{-1}=6 \times 10^{26}\ \text{W}\)
\(\Rightarrow C\)
A company produces soft drinks in aluminium cans.
The company sources empty cans from an external supplier, who claims that the mass of aluminium in each can is normally distributed with a mean of 15 grams and a standard deviation of 0.25 grams.
A random sample of 64 empty cans was taken and the mean mass of the sample was found to be 14.94 grams.
Uncertain about the supplier's claim, the company will conduct a one-tailed test at the 5% level of significance. Assume that the standard deviation for the test is 0.25 grams.
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The equipment used to package the soft drink weighs each can after the can is filled. It is known from past experience that the masses of cans filled with the soft drink produced by the company are normally distributed with a mean of 406 grams and a standard deviation of 5 grams.
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a. \(H_0: \mu=15, \quad H_1: \mu<15\)
b. \(p=0.027\)
c. \(\text{Since \(\ p<0.05\), claim is not supported.}\)
d. \(a=14.95\)
e. \(\text{Pr}(-3<D<3)=0.329\)
a. \(H_0: \mu=15, \quad H_1: \mu<15\)
b. \(\mu=15, \ \ \sigma=0.25\)
\(\bar{x}=14.94, \ \sigma_{\bar{x}}=\dfrac{0.25}{\sqrt{64}}=0.03125\)
\(\text{By CAS:}\)
\(p=\text{Pr}\left(\bar{X}<14.94 \mid \mu=15\right)=0.027\ \text {(3 d.p.)}\)
c. \(\text{Since \(\ p<0.05\), claim is not supported.}\)
\(\text{Evidence is against \(H_0\) at the \(5 \%\) level.}\)
d. \(\text{Pr}\left(\bar{X}<a \mid \mu=15\right)>0.05\)
\(\text{Pr}\left(Z<\dfrac{a-15}{0.03125}\right)>0.05\)
\(\text{By CAS:}\ \ a=14.95\ \text{(2 d.p.)}\)
e. \(\text{Let}\ \ M=\ \text{mass of one can}\)
\(M \sim N\left(406,5^2\right)\)
\(E\left(M_1\right)=E\left(M_2\right)=\mu=406\)
\(\text {Let}\ \ D=M_1-M_2\)
\(E(D)=406-406=0\)
\(\text{Var}(D)=1^2 \times \text{Var}\left(M_1\right)+(-1)^2 \times \text{Var}\left(M_2\right)=50\)
\(\sigma_D=\sqrt{50}\)
\(D \sim N\left(0,(\sqrt{50})^2\right)\)
\(\text{By CAS: Pr\((-3<D<3)=0.329\) (3 d.p.) }\)
A student is playing minigolf on a day when there is a very strong wind, which affects the path of the ball. The student hits the ball so that at time \(t=0\) seconds it passes through a fixed origin \(O\). The student aims to hit the ball into a hole that is 7 m from \(O\). When the ball passes through \(O\), its path makes an angle of \(\theta\) degrees to the forward direction, as shown in the diagram below.
The path of the ball \(t\) seconds after passing through \(O\) is given by
\(\underset{\sim}{\text{r}}(t)=\dfrac{1}{2} \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{\text{i}}+2 t \underset{\sim}{\text{j}}\) for \(t \in[0,5]\)
where \(\underset{\sim}{i}\) is a unit vector to the right, perpendicular to the forward direction, \(\underset{\sim}{j}\) is a unit vector in the forward direction and displacement components are measured in metres.
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a. \(\theta =11.1^{\circ}\)
b.i. \(2.04\ \text{ms}^{-1}\)
b.ii. \(\text{Minimum speed} =2 \text{ ms} ^{-1}\)
c. \(\text {Minimum }\abs{d}=0.188\ \text{m}\)
d. \(8.077\ \text{m}\)
a. \(r(t)=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 t \underset{\sim}{j}\)
\(\dot{r}(t)=\dfrac{\pi}{8}\, \cos \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 \underset{\sim}{j}\)
\(\dot{r}(0)=\dfrac{\pi}{8}\,\underset{\sim}{i}+2\underset{\sim}{j}\)
\begin{aligned}
\tan (90-\theta) & =\dfrac{2}{\frac{\pi}{8}} \\
90-\theta & =\tan ^{-1}\left(\dfrac{16}{\pi}\right)=78.9^{\circ} \\
\theta & =11.1^{\circ}
\end{aligned}
b.i. | \(\text{Speed}\) | \(=\abs{\dot{r}(0)}\) |
\(=\sqrt{\left(\dfrac{\pi}{8}\right)^2+2^2}\) | ||
\(=2.04\ \text{ms}^{-1}\ \text{(2 d.p.)}\) |
b.ii. \(\abs{\dot{r}}=\sqrt{\left(\dfrac{\pi}{8}\right)^2 \times \cos ^2\left(\dfrac{\pi t}{4}\right)+4}\)
\(\abs{\dot{r}}\text { is a minimum when}\ \ \cos ^2\left(\dfrac{\pi t}{4}\right)=0\)
\(\Rightarrow t=2\)
\(\therefore\ \text{Minimum speed} =\sqrt{4}=2 \text{ ms} ^{-1}\)
c. \(\text{Ball position}\ \Rightarrow \ \underset{\sim}{r}=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right)\underset{\sim}{i}+2 \, t\underset{\sim}{j}\)
\(\text{Hole position}\ \Rightarrow \ \underset{\sim}{h}=0 \underset{\sim}{i}+7\underset{\sim}{j}\)
\(\underset{\sim}{d}=\underset{\sim}{r}-\underset{\sim}{h}=\dfrac{1}{2}\, \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+(2 t-7)\underset{\sim}{j}\)
\(\abs{\underset{\sim}{d}}=\sqrt{\left(\dfrac{1}{2}\right)^2 \sin ^2\left(\dfrac{\pi t}{4}\right)+(2 t-7)^2}\)
\(\text {Minimum }\abs{d}=0.188\text{ m (3 d.p.)}\)
d. | \(\text{Distance}\) | \(=\displaystyle \int_0^4\left(\frac{\pi}{8}\, \cos \left(\frac{\pi t}{4}\right)\right)^2+4\, dt\) |
\(=8.077\ \text{m (3 d.p.)}\) |
Write balanced chemical equations for each of the following reactions (states of matter are not required).
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a. \(\ce{LiOH + HCl -> LiCl + H2O}\)
b. \(\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\)
a. Acid-Base
\[\ce{LiOH + HCl -> LiCl + H2O}\]
b. Acid-carbonate
\[\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\]
Write balanced chemical equations for each of the following reactions (states of matter are not required).
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a. \(\ce{Zn + H2SO4 -> ZnSO4 + H2}\)
b. \(\ce{CaCO3 -> CaO + CO2}\)
c. \(\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\)
a. Active metal and acid
\[\ce{Zn + H2SO4 -> ZnSO4 + H2}\]
b. Decomposition
\[\ce{CaCO3 -> CaO + CO2}\]
c. Combustion
\[\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\]
Write balanced chemical equations for each of the following reactions (states of matter are not required).
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a. \(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)
b. \(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)
a. Acid-base
\(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)
b. Acid-carbonate
\(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)
Write balanced chemical equations for each of the following reactions (states of matter are not required).
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a. \(\ce{Mg + 2HCl -> MgCl2 + H2}\)
b. \(\ce{2NH3 -> N2 + 3H2}\)
c. \(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)
a. Reaction between active metal and acid
\(\ce{Mg + 2HCl -> MgCl2 + H2}\)
b. Decomposition
\(\ce{2NH3 -> N2 + 3H2}\)
c. Combustion
\(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)
A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.
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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
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a. \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
b. \(2.06 \text{ g}\)
c. Possible reasons for the higher mass:
→ Presence of excess solution in the filter paper or incomplete drying.
→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
Experimental adjustment to eliminate error:
→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
→ Dry precipitate completely in the incubator before weighing.
a. \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
b. \(\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}\)
\(\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}\)
\(\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}\)
\(\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}\)
\(\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}\)
\(\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}\)
c. Possible reasons for the higher mass:
→ Presence of excess solution in the filter paper or incomplete drying.
→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
Experimental adjustment to eliminate error:
→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
→ Dry precipitate completely in the incubator before weighing.
A student stirs 2.45 g of copper \(\text{(II)}\) nitrate powder into 200.0 mL of 1.25 mol L\(^{-1}\) sodium carbonate solution until it is fully dissolved. A reaction occurs and a precipitate appears.
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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
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a. \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)
b. \(1.61 \text{ g}\)
c. Possible reasons for the higher mass:
→ Presence of excess solution in the filter paper or incomplete drying.
→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
Experimental adjustment to eliminate error:
→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
→ Dry precipitate completely in the incubator before weighing.
a. \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)
b. \(\ce{MM(Cu(NO3)2) = 63.55 + 2 \times (14.01 + 16.00 \times 3) = 187.57}\)
\(\ce{n(Cu(NO3)2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.45}{187.57} = 0.01306 \text{ mol}}\)
\(\ce{n(Na2CO3) = c \times V = 1.25 \times 0.200 = 0.250 \text{ mol}}\)
\(\Rightarrow \ce{Cu(NO3)2} \text{ is the limiting reagent}\)
\(\ce{n(CuCO3) = n(Cu(NO3)2) = 0.01306 \text{ mol}}\)
\(\ce{m(CuCO3) = n \times MM = 0.01306 \times (63.55 + 12.01 + 16.00 \times 3) = 1.61 \text{ g}}\)
c. Possible reasons for the higher mass:
→ Presence of excess solution in the filter paper or incomplete drying.
→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
Experimental adjustment to eliminate error:
→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
→ Dry precipitate completely in the incubator before weighing.
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a. Determine the empirical formula of each compound.
Butyraldehyde \(\ce{(C4H8O)}\): Empirical formula = \(\ce{C4H8O}\)
Lactic acid \(\ce{(C3H6O3)}\): Empirical formula = \(\ce{CH2O}\)
Fructose \(\ce{(C6H12O6)}\): Empirical formula = \(\ce{CH2O}\)
→ Butyraldehyde and lactic acid have the same empirical formula of \(\ce{CH2O}\).
b. \(\ce{C16H20O8}\)
a. Determine the empirical formula of each compound.
Butyraldehyde \(\ce{(C4H8O)}\): Empirical formula = \(\ce{C4H8O}\)
Lactic acid \(\ce{(C3H6O3)}\): Empirical formula = \(\ce{CH2O}\)
Fructose \(\ce{(C6H12O6)}\): Empirical formula = \(\ce{CH2O}\)
→ Butyraldehyde and lactic acid have the same empirical formula of \(\ce{CH2O}\).
b. Calculate the molar mass of the empirical formula \(\ce{C4H5O2}\):
\(\text{Molar mass}\ \ce{(C4H5O2)} = 4 \times 12.01 + 5 \times 1.008 + 2 \times 16.00 = 85.08\ \text{g/mol}\)
Ratio of the molar mass of the compound to the molar mass of the empirical formula:
\(\text{Ratio} = \dfrac{\text{Molar mass}}{\text{Empirical formula mass}} = \dfrac{340.32\ \text{g mol}^{-1}}{85.08\ \text{g mol}^{-1}} =4\)
Multiply the subscripts in the empirical formula by 4 to get the molecular formula:
\(\text{Molecular formula} = \ce{(C4H5O2)} \times 4 = \ce{C16H20O8}\)
→ Thus, the molecular formula of the compound is \(\ce{C16H20O8}\).
During a laboratory experiment, 32.0 grams of oxygen gas \(\ce{(O2)}\) is required. Calculate the number of molecules of oxygen gas needed. (2 marks)
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\(6.022 \times 10^{23}\) molecules.
Calculate the number of moles of oxygen gas (\(\ce{O2}\)):
\(\ce{n(O2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{32.0\ \text{g}}{32.00\ \text{g mol}^{-1}} = 1.00\ \text{mol}}\)
Use Avogadro’s number to find the number of molecules:
\(\ce{N(O2) = n \times N_A = 1.00 \, mol \times 6.022 \times 10^{23} \, mol^{-1} = 6.022 \times 10^{23} \, molecules}\)
→ Oxygen gas required = \(6.022 \times 10^{23}\) molecules.
Ammonia \(\ce{(NH3)}\) is synthesized from nitrogen \(\ce{(N2)}\) and hydrogen \(\ce{(H2)}\) according to the following equation:
\(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\)
If 1.50 moles of nitrogen gas react completely with hydrogen gas, calculate the mass of ammonia produced. (2 marks)
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\(\ce{m(NH3) = 51.0\ \text{g}}\)
Calculate the number of moles of ammonia \(\ce{(NH3)}\) produced:
\(\ce{n(NH3) = 2 \times n(N2) = 2 \times 1.50 \, mol = 3.00 \, mol}\)
Calculate the mass of ammonia produced:
\(\ce{m(NH3) = n \times MM = 3.00 \, mol \times 17.03\ \text{g mol}^{-1} = 51.09\ \text{g}}\)
→ The mass of ammonia produced is 51.0 grams.
The function \(g\) is defined as follows.
\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)
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Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).
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a.
b.i | \(g(x)\) | \(=3\log_{e}x-x\) |
\(g(1)\) | \(=3\log_{e}1-1=-1\) | |
\(g^{\prime}(x)\) | \(=\dfrac{3}{x}-1\) | |
\(g^{\prime}(1)\) | \(=\dfrac{3}{1}-1=2\) |
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)
\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)
b.ii
c. \(\text{Newton’s Method}\)
\(x_1\) | \(=x_0-\dfrac{g(x)}{g'(x)}\) |
\(=1-\left(\dfrac{-1}{2}\right)\) | |
\(=\dfrac{3}{2}=1.5\) |
\(\text{Using CAS:}\)
d. \(\text{Using CAS}\)
\(x\text{-intercept}:\ x=1.85718\)
\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)
e.i. \(\text{Using CAS}\)
\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) | \(=1.5\) |
\(k>1\ \therefore\ \ k\) | \(=2.397\) |
e.ii
During a laboratory experiment, a gas is collected in a sealed syringe. Initially, the gas has a volume of 5.0 litres and a pressure of 1.0 atmosphere.
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a. \(1.67\ \text{atm}\)
b. \(1.5\ \text{L}\)
c. Potential experimental errors could include:
→ Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.
→ Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.
→ Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.
a. Using Boyle’s Law \( P_1V_1 = P_2V_2 \):
\( P_1 = 1.0 \, \text{atm}, \quad V_1 = 5.0 \, \text{L}, \quad V_2 = 3.0 \, \text{L} \)
\(P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.0 \times 5.0}{3.0} = 1.67 \, \text{atm} \)
b. To find the new volume when pressure doubles:
\( P_3 = 2 \times P_2 = 2 \times 1.67 \, \text{atm} = 3.34 \, \text{atm}\)
\( P_1V_1 = P_3V_3 \ \ \Rightarrow \ \ V_3 = \dfrac{P_1 \times V_1}{P_3} = \dfrac{1.0 \times 5.0 }{3.34} \approx 1.5 \, \text{L}\)
c. Potential experimental errors could include:
→ Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.
→ Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.
→ Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.
A cylinder equipped with a movable piston contains an inert gas. The initial pressure of the gas is 1.5 atmospheres, and the cylinder's initial volume is 4.0 litres.
The piston is then adjusted to change the volume of the cylinder.
Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume is adjusted to 2.0 litres. (2 marks)
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\(3.0\ \text{atm}\)
\(P_1 = 1.5\ \text{atm},\ \ V_1 = 4.0\ \text{L},\ \ V_2 = 2.0\ \text{L} \)
Boyle’s Law \(\Rightarrow \ P_1V_1 = P_2V_2 \).
\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.5 \, \text{atm} \times 4.0 \, \text{L}}{2.0 \, \text{L}} = 3.0 \, \text{atm}\)
A cylinder equipped with a movable piston contains an inert gas. The initial pressure of the gas is 1.35 atmospheres, and the cylinder's initial volume is 5.0 litres.
The piston is then adjusted to change the volume of the cylinder.
Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume of the cylinder is adjusted to 8.0 litres. (2 marks)
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\(0.84\ \text{atm}\)
\(V_1 = 5.0\ \text{L},\ \ V_2 = 8.0\ \text{L},\ \ P_1=1.35\ \text{atm} \)
Boyle’s Law \(\Rightarrow \ P_1V_1 = P_2V_2 \).
\(P_2 = \dfrac{P_1 \times V_1}{V_{2}} = \dfrac{1.35\ \text{atm} \times 5.0\ \text{L}}{8.0\ \text{L}} = 0.84\ \text{atm}\)
A mixture of sand and salt was provided to a group of students for them to determine its percentage composition by mass.
They added water to the sample before using filtration and evaporation to separate the components.
During the evaporation step, the students noticed white powder ‘spitting’ out of the basin onto the bench, so they turned off the Bunsen burner and allowed the remaining water to evaporate overnight.
After filtering, they allowed the filter paper to dry overnight before weighing. An electronic balance was used to measure the mass of each component to two decimal places.
The results were recorded as shown:
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a. % Sand = 70.88%
% Salt = 19.71%
b. → The calculations show that the percentages do not add up to 100.
→ Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.
→ Hence, experiment is not valid.
a. | \(\text{Sand mass}\ \) | \(\text{ = Mass of dried filter paper – Mass of filter paper}\) |
\(= 11.95-0.80 = 11.15\ \text{g}\) |
\(\text{Salt mass}\) | \(\text{ = Mass of dried filter paper – Mass of filter paper}\) | |
\(= 36.60-33.50 = 3.10\ \text{g}\) |
\(\text{% sand} = \left(\dfrac{\text{Mass of sand}}{\text{Mass of original mixture}}\right) \times 100= \left(\dfrac{11.15 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 70.88\%\)
\(\text{% salt} = \left(\dfrac{\text{Mass of salt}}{\text{Mass of original mixture}}\right) \times 100 = \left(\dfrac{3.10 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 19.71\%\)
b. → The calculations show that the percentages do not add up to 100.
→ Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.
→ Hence, experiment is not valid.
A particle moves in a straight line so that its distance, \(x\) metres, from a fixed origin \(O\) after time \(t\) seconds is given by the differential equation \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\), where \(x=0\) when \(t=0\).
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Two seconds after the first particle passed through \(O\), a second particle passes through \(O\).
Its distance \(x\) metres from \(O, t\) seconds after the first particle passed through \(O\), is given by \(x=\log _e\left(\tan ^{-1}(3 t-6)+1\right).\)
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a.i. \(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
a.ii. \(x=\log _e\left(\tan ^{-1}(2 t)+1\right) \)
b.i. \(\text{Asymptote: }\ x=\log _e\left(\dfrac{\pi}{2}+1\right)\)
b.ii.
c. \(0.02\)
d. \(\text {Particle positions at}\ \ t=6:\)
\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)
\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
e. \(\dfrac{v_1}{v_2}=\dfrac{2}{3}\)
a.i. \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\)
\(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
a.ii. \(e^x=\tan ^{-1}(2 t)+c\)
\(\text {When}\ \ t=0, \ x=0 \ \Rightarrow \ c=1\)
\begin{aligned}
e^x& =\tan ^{-1}(2 t)+1 \\
x&=\log _e\left(\tan ^{-1}(2 t)+1\right)
\end{aligned}
b.i. \(\text {As}\ \ t \rightarrow \infty, \ \tan ^{-1}(2 t) \rightarrow \dfrac{\pi}{2}\)
\(x \rightarrow \log _e\left(\dfrac{\pi}{2}+1\right)\)
\(\therefore \text{Asymptote: }\ x=\log _e\left(\dfrac{\pi}{2}+1\right)\)
b.ii. \(\log _e\left(\dfrac{\pi}{2}+1\right) \approx 0.944\)
\(\text{When}\ \ t=10 :\)
\(x=\log _e\left( \tan ^{-1}(20)+1\right)=0.92\ \text{(2 d.p.)}\)
c. \(\text {At}\ \ t=3, x=\log _e\left(\tan ^{-1}(6)+1\right)\)
\(\text {Substitute } t \text { and } x \text { into } \dfrac{d x}{d t}:\)
\(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}=0.02\, \text{ms} ^{-1}\ \text{(2 d.p.)}\)
d. \(\text {Particle positions at}\ \ t=6:\)
\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)
\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
e. \(e^x=e^{\log _e\left(\tan ^{-1}(12)+1\right)}=\tan ^{-1}(12)+1\)
\(\text {At}\ \ t=6:\)
\(\dfrac{d x_1}{d t}=\dfrac{2}{1+4 t^2} \times \dfrac{1}{\tan ^{-1}(2 t)+1}=\dfrac{2}{145\left(\tan ^{-1}(12)+1\right)}\)
\(\dfrac{d x^2}{d t}=\dfrac{3}{1+(3 t-6)^2} \times \dfrac{1}{\tan ^{-1}(3 t-6)-1}=\dfrac{3}{145\left(\tan ^{-1}(12)+1\right)}\)
\(\therefore \dfrac{v_1}{v_2}=\dfrac{2}{3}\)
Discuss the reasons why Neon, similar to other noble gases, does not easily form chemical bonds with other elements.
Explain Neon's electron configuration and how this relates to its position on the periodic table and its chemical characteristics. (3 marks)
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→ Neon is located in Group 18 of the periodic table and has an atomic number of 10.
→ Its electron configuration is 1s² 2s² 2p⁶. This configuration shows that Neon has a completely filled outer shell, a characteristic of noble gases, which contributes to its chemical inertness.
→ As a member of Group 18, Neon shares this fully filled valence shell characteristic with other noble gases, resulting in high stability and very low reactivity.
→ Neon is located in Group 18 of the periodic table and has an atomic number of 10.
→ Its electron configuration is 1s² 2s² 2p⁶. This configuration shows that Neon has a completely filled outer shell, a characteristic of noble gases, which contributes to its chemical inertness.
→ As a member of Group 18, Neon shares this fully filled valence shell characteristic with other noble gases, resulting in high stability and very low reactivity.
You are given the task to separate the components of two mixtures: a saltwater solution and a mixture of sand and iron filings.
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a. Separation technique:
→ Evaporation is a suitable technique because water has a much lower boiling point than salt.
→ By heating the solution, the water evaporates, leaving salt crystals behind.
→ Further distillation could be used to collect and condense the evaporated water.
b. Physical property:
→ Iron fillings are magnetic and hence can be separated using a magnet.
→ The mixture is heterogeneous.
c. Safety precaution:
→ For evaporation, wear heat-resistant gloves when handling hot equipment like the tripod or beaker to avoid burns to the hands.
a. Separation technique:
→ Evaporation is a suitable technique because water has a much lower boiling point than salt.
→ By heating the solution, the water evaporates, leaving salt crystals behind.
→ Further distillation could be used to collect and condense the evaporated water.
b. Physical property:
→ Iron fillings are magnetic and hence can be separated using a magnet.
→ The mixture is heterogeneous.
c. Safety precaution:
→ For evaporation, wear heat-resistant gloves when handling hot equipment like the tripod or beaker to avoid burns to the hands.
The probability density function \(f\) of a random variable \(X\) is given by
\(f(x)= \begin{cases}\dfrac{x+1}{20} & 0 \leq x<4 \\ \dfrac{36-5 x}{64} & 4 \leq x \leq 7.2 \\ 0 & \text {elsewhere}\end{cases}\)
The value of \(a\) such that \(\text{Pr}(X \leq a)=\dfrac{5}{8}\) is
\(\displaystyle \int_{0}^{4}\dfrac{x+1}{20}\,dx =\dfrac{3}{5}<\dfrac{5}{8}\)
\(\therefore\ \text{Pr}(0\leq X<4)+\text{Pr}(4\leq X<a)=\dfrac{5}{8}\)
\(\rightarrow\quad\) | \(\dfrac{3}{5}+\displaystyle \int_{4}^{a}\dfrac{36-5x}{64}\,dx\) | \(=\dfrac{5}{8}\) |
\(\displaystyle \int_{4}^{a}\dfrac{36-5x}{64}\,dx\) | \(=\dfrac{5}{8}-\dfrac{3}{5}=\dfrac{1}{40}\) | |
\(\therefore\ \text{from CAS:}\quad \ a\) |
\(=\dfrac{-4\Big(\sqrt{15}-9\Big)}{5}\) |
|
\(=\dfrac{36-4\sqrt{15}}{5}\) |
\(\text{Using CAS:}\)
\(\Rightarrow C\)
The area between the curve \(\displaystyle y=\frac{1}{27}(x-3)^2(x+3)^2+1\) and the \(x\)-axis on the interval \(x \in[0,4]\) has been approximated using the trapezium rule, as shown in the graph below.
Using the trapezium rule, the approximate area calculated is equal to
\(B\)
\(\text{Area each trapezium}=\dfrac{h}{2}(f(a)+f(b))\)
\(A\) | \(=\Bigg(\dfrac{1}{2}(f(0)+f(1))+\dfrac{1}{2}(f(1)+f(2))+\dfrac{1}{2}(f(2)+f(3))+\dfrac{1}{2}(f(3)+f(4)\Bigg)\) |
\(=\dfrac{1}{2}\Bigg(f(0)+2f(1)+2f(2)+2f(3)+f(4)\Bigg)\) | |
\(=\dfrac{1}{2}\Bigg(4+2\times\dfrac{91}{27}+2\times\dfrac{52}{27}+2\times 1+\dfrac{76}{27}\Bigg)\) | |
\(=\dfrac{1}{2}\Bigg(4+\dfrac{182}{27}+\dfrac{104}{27}+2+\dfrac{76}{27}\Bigg)\) |
\(\Rightarrow B\)
\begin{aligned}
& x-2 y=3 \\
& 2 y-z=4
\end{aligned}
Which one of the following correctly describes the general solution to the system of linear equations given above?
Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.
Use the following table of values for `f` and `f^{\prime}`.
`\quad x \quad` | `\quad\quad 1/2\quad\quad` | `\quad\quad(sqrt{2})/2\quad\quad` | `\quad\quad(sqrt{3})/2\quad\quad` |
`f(x)` | `-2` | `5` | `3` |
`\quad\quad f^{prime}(x)\quad\quad` | `7` | `0` | `1/9` |
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The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.
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a. `3`
b. `1/9`
c. `y=1/9x+3-pi/54`
d. `-48/pi`
e. ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`
a. `g(pi/6)` |
`= f(sin(pi/3))` | |
`= f(sqrt3/2)` | ||
`= 3` |
b. `g\ ^{prime}(x)` | `= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))` | |
`g\ ^{prime}(pi/6)` | `= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)` | |
`= 1/9` |
c. `m = 1/9` and `g(pi/6) = 3`
`y – y_1` | `= m(x-x_1)` | |
`y – 3` | `= 1/9(x-pi/6)` | |
`y` | `= 1/9x + 3-pi/54` |
d. The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`
Average | `= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x` | |
`=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}` | ||
`= 24/pi \cdot(f(sqrt3/2)-f(sqrt2/2))` | ||
`= 24/pi (3-5) = -48/pi` |
e. `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`
`:.\ 2 \cos (2 x) = 0\ ….(1)` or ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`
(1): ` 2 \cos (2 x)` | `= 0` | `x \in[0, \pi]` |
`\cos (2 x)` | `= 0` | `2 x \in[0,2 \pi]` |
`2x` | `= pi/2 , (3pi)/2` | |
`x` | `= pi/4 , (3pi)/4` | |
(2): ` f^{\prime}(\sin (2 x)) ` | `= sqrt2/2` | |
`2x` | `= pi/4 , (3pi)/4` | |
`x` | `= pi/8 , (3pi)/8` |
`:. \ x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`
Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`
Part of the graph of `y=f(x)` is shown below.
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a. | `R` |
b.i | `f^{\prime}(0)=4` |
b.ii | `\left(-\frac{1}{2}, \frac{1}{2}\right)` |
c. | `0` |
d. | `x \in \mathbb{R}` |
e.i | ` k > 4` |
e.ii | No bounded area for `0<k \leq 4` |
a. `R` is the range.
b.i | `f(x)` |
`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)` | |
`f^{\prime}(x)` | `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}` | ||
`= \frac{2}{2 x+1}-\frac{2}{2 x-1}` | |||
`f^{\prime}(0)` | `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}` | ||
`= 4` |
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `f(x)+f(-x)`
`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)` | |
`= 0` |
d. To find the inverse swap `x` and `y` in `y=f(x)`
`x` | `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)` | |
`x` | `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)` | |
`e^x` | `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}` | |
`y+\frac{1}{2}` | `= e^x\left(-y+\frac{1}{2}\right)` | |
`y+\frac{1}{2}` | `= -e^x y+\frac{e^x}{2}` | |
`y\left(e^x+1\right)` | `= \frac{e^x-1}{2}` | |
`:.\ f^(-1)(x)` | `= \frac{e^x-1}{2(e^x + 1)}` |
`:.` Domain: `x \in \mathbb{R}`
e.i The vertical dilation factor of `f(x)` is `1/k`
For `A(k)>=0` , `h^{\prime}(0)<1`
`\frac{1}{k}(4)<1` [Using CAS]
`:.\ k > 4`
e.ii When `h \geq h^{-1}` for `x>0` (or `h \leq h^{-1}` for `x<0`) there is no bounded area.
`:.` There will be no bounded area for `0<k \leq 4`.
Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.
Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.
Mika flips the coin five times.
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The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
\(f(h)=\begin{cases} ah^2+bh+c &\ \ 1.5\leq h\leq 3 \\ \\ 0 &\ \ \text{elsewhere} \\ \end{cases}\)
where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.
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a. i. `frac{1}{32}` ii. `frac{13}{16}` iii. `0.806` (3 d.p.)
a. iv `text{E}(X)=5/2, text{sd}(X)=\frac{\sqrt{5}}{2}`
b. i. `1`
b. ii. `a=-frac{4}{5}, b=frac{17}{5}, c=-frac{167}{60}`
b. iii. `r=-1, s=3`
c. i. `text{Discrete}` ii. `(0.208, 0.592)` iii. `n=100`
a.i `X ~ text{Bi}(5 , frac{1}{2})`
`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
a.ii By CAS: `text{binomCdf}(5,0.5,2,5)` `0.8125`
`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`
a.iii `\text{Pr}(X \geq 2 | X<5)`
`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
`= 0.806452 ~~ 0.806` (3 decimal places)
a.iv `X ~ text{Bi}(5 , frac{1}{2})`
`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`
`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`
b.i `\int_{1.5}^3 f(h) d h = 1`
b.ii By CAS:
`f(h):= a\·\h^2 + b\·\h +c`
`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`
`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\ c = =-2.78 \dot{3} = frac{-167}{60}`
b.iii `h + d = 3`
`:.\ f(h) = f(3 – d) = f(- d + 3)`
`:.\ r = – 1 ` and ` s = 3`
c.i `\hat{p}` is discrete.
The number of coin flips must be zero or a positive integer so `\hat{p}` is countable and therefore discrete.
c.ii `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`
`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`
`\approx(0.208\ ,0.592)`
c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.
`:.\ \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`
`:.\ n = 100`
She would need to flip the coin 100 times
Show that the parabola \(2x^2-kx+k-2\) has at least one real root. (3 marks)
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\(2x^2-kx+k-2=0\)
\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)
\(\text{Real roots:}\ \ \Delta \geqslant 0\)
\(k^2-8k+16\) | \(\geqslant 0\) | |
\((x-4)^2\) | \(\geqslant 0\) |
\(\therefore\ \text{At least one root exists for all}\ k\)
\(2x^2-kx+k-2=0\)
\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)
\(\text{Real roots:}\ \ \Delta \geqslant 0\)
\(k^2-8k+16\) | \(\geqslant 0\) | |
\((x-4)^2\) | \(\geqslant 0\) |
\(\therefore\ \text{At least one root exists for all}\ k\)
Silver oxide, when heated, decomposes according to the following chemical equation:
\(\ce{2Ag2O -> 4Ag + O2}\)
231.74 grams of silver oxide (\(\ce{Ag2O}\)) is heated, and the metallic silver (\(\ce{Ag}\)) residue is weighed and recorded in the table below.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{231.74 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{216.0 grams} \\
\hline
\end{array}
Calculate the percentage of oxygen (\(\ce{O2}\)), by weight, released from the silver oxide during this process. (2 marks)
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\(\%\ce{O2} = 6.8\%\)
\(\text{Initial mass of}\ \ce{Ag2O} = 231.74\ \text{grams}\)
\(\text{Mass of}\ \ce{Ag} = 216.0\ \text{grams}\)
\(\text{Mass of}\ \ce{O2} = 231.74-216.0 = 15.74\ \text{grams}\)
\(\%\ce{O2}\ \text{released}\ = \left(\dfrac{15.74}{231.74} \times 100\%\right) = 6.8\%\)
Two complex numbers \(u\) and \(v\) are given by \(u=a+i\) and \(v=b-\sqrt{2}i\), where \(a, b \in R\).
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a.i. \(u v=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6}) i\)
\begin{aligned}
(a+i)(b-\sqrt{2} i) & =a b-\sqrt{2} a i+b i+\sqrt{2} \\
& =a b+\sqrt{2}+(b-\sqrt{2}a) i
\end{aligned}
\(\text {Equating co-efficients: }\)
\(ab=\sqrt{6} \ \Rightarrow \ b=\dfrac{\sqrt{6}}{a}\ \cdots\ \text {(1)}\)
\(b-\sqrt{2} a=\sqrt{2}-\sqrt{6}\ \cdots\ \text {(2)}\)
\(\text{Substitute (1) into (2):}\)
\(\dfrac{\sqrt{6}}{a}-\sqrt{2}a=\sqrt{2}-\sqrt{6}\)
\(\sqrt{6}-\sqrt{2} a^2=(\sqrt{2}-\sqrt{6}) a\)
\(\sqrt{2} a^2+(\sqrt{2}-\sqrt{6}) a-\sqrt{6}\) | \(=0\) | |
\(a^2+(1-\sqrt{3}) a-\sqrt{3}\) | \(=0\) |
a.ii. \(\text{Solve } a^2+(1-\sqrt{3}) a-\sqrt{3}=0 \ \ \text{for}\ a :\)
\(a=\sqrt{3} \ \text{ or } -1\)
\(\therefore \text{ Other solution: } a=-1, \ b=-\sqrt{6}\)
c. |
\begin{aligned} \theta & =\frac{\operatorname{Arg}(u)-\operatorname{Arg}(v)}{2} \\ & =\frac{1}{2}\left(\frac{\pi}{6}-\frac{\pi}{4}\right) \\ & =-\frac{\pi}{24} \end{aligned} |
d. \(\text {Sector angle (centre) }=\dfrac{\pi}{6}+\dfrac{\pi}{4}=\dfrac{5 \pi}{12}\)
\(\text {Area of sector }=\dfrac{\frac{5 \pi}{12}}{2 \pi} \times \pi \times 2^2=\dfrac{5 \pi}{6}\)
\begin{aligned}
\text {Area of segment } & =\frac{5 \pi}{6}-\dfrac{1}{2} \times 2 \times 2 \times \sin \left(\dfrac{5 \pi}{6}\right) \\
& \approx 0.69\ \text{u} ^2
\end{aligned}
Consider the family of functions \(f\) with rule \(f(x)=\dfrac{x^2}{x-k}\), where \(k \in R \backslash\{0\}\).
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a. \(\text {Asymptotes: } x=1,\ y=x+1\)
b.
c.i. \(\text {Asymptotes: } x=k,\ y=x+k\)
c.ii. \(\text {Distance }=2 \sqrt{5}|k|\)
d.i. \(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
d.ii. \(V=51.42\ \text{u}^3 \)
a. \(\text {When } k=1 :\)
\(f(x)=\dfrac{x^2}{x-1}=\dfrac{(x+1)(x-1)+1}{(x-1)}=x+1+\dfrac{1}{x-1}\)
\(\text {Asymptotes: } x=1,\ y=x+1\)
b.
c.i. \(f(x)=\dfrac{x^2}{x-k}=\dfrac{(x+k)(x-k)+k^2}{x-k}=x+k+\dfrac{k^2}{x-k}\)
\(\text {Using part a.}\)
\(\text {Asymptotes: } x=k,\ y=x+k\)
c.ii. \(f^{\prime}(x)=1-\left(\dfrac{k}{x-k}\right)^2\)
\(\text {TP’s when } f^{\prime}(x)=0 \text { (by CAS):}\)
\(\Rightarrow(2 k, 4 k),(0,0)\)
\(\text {Distance }\displaystyle=\sqrt{(2 k-0)^2+(4 k-0)^2}=\sqrt{20 k^2}=2 \sqrt{5}|k|\)
d.i \(\text {Solve for intersection of graphs (by CAS):}\)
\(\displaystyle x+3=\left|\frac{x^2}{x-1}\right|\)
\(\displaystyle \Rightarrow x=\frac{3}{2}, x=\frac{-1 \pm \sqrt{7}}{2}\)
\(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
d.ii. \(V=51.42\ \text{u}^3 \text{ (by CAS) }\)
Consider the vectors \(\underset{\sim}{\text{u}}(x)=-\text{cosec}(x) \underset{\sim}{\text{i}}+\sqrt{3} \underset{\sim}{\text{j}}\) and \(\underset{\sim}{\text{v}}(x)=\text{cos}(x) \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\).
If \(\underset{\sim}{\text{u}}(x)\) is perpendicular to \(\underset{\sim}{\text{v}}(x)\), then possible values for \(x\) are
\(A\)
\(\underset{\sim}{u} \perp \underset{\sim}{v} \ \Rightarrow \ \underset{\sim}{u} \cdot \underset{\sim}{v}=0\)
\(-\text{cosec}(x) \cos (x)+\sqrt{3}=0\)
\begin{aligned}
-\dfrac{\cos (x)}{\sin (x)}+\sqrt{3} & =0 \\
\tan (x) & =\dfrac{1}{\sqrt{3}}
\end{aligned}
\(x=\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\)
\(\Rightarrow A\)
Consider the vectors \(\underset{\sim}{\text{u}}(x)=-\text{cosec}(x) \underset{\sim}{\text{i}}+\sqrt{3} \underset{\sim}{\text{j}}\) and \(\underset{\sim}{\text{v}}(x)=\text{cos}(x) \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\).
If \(\underset{\sim}{\text{u}}(x)\) is perpendicular to \(\underset{\sim}{\text{v}}(x)\), then possible values for \(x\) are
\(A\)
\(\underset{\sim}{u} \perp \underset{\sim}{v} \ \Rightarrow \ \underset{\sim}{u} \cdot \underset{\sim}{v}=0\)
\(-\text{cosec}(x) \cos (x)+\sqrt{3}=0\)
\begin{aligned}
-\dfrac{\cos (x)}{\sin (x)}+\sqrt{3} & =0 \\
\tan (x) & =\dfrac{1}{\sqrt{3}}
\end{aligned}
\(\text {Solve by (CAS)}: \ x=\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\)
\(\Rightarrow A\)
Consider the vectors \(\underset{\sim}{\text{a}}=2 \underset{\sim}{\text{i}}-3 \underset{\sim}{\text{j}}+p \underset{\sim}{\text{k}}, \ \underset{\sim}{\text{b}}=\underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}-q \underset{\sim}{\text{k}}\) and \(\underset{\sim}{\text{c}}=-3 \underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}+5 \underset{\sim}{\text{k}}\), where \(p\) and \(q\) are real numbers.
If these vectors are linearly dependent, then
\(A\)
\(\text{Linearly dependent when}\ \underset{\sim}{c}=m\underset{\sim}{a}+n\underset{\sim}{b}\)
\(\underset{\sim}{c}=-3 \underset{\sim}{i}+2 \underset{\sim}{j}+\underset{\sim}{k}\)
\(m\underset{\sim}{a}+n \underset{\sim}{b}=(2 m+n) \underset{\sim}{i}-(3 m-2 n) \underset{\sim}{j}+(m p-q n) \underset{\sim}{k}\)
\(\text{Equating co-efficients:}\)
\(2 m+n=-3 \ \ldots\ (1)\)
\(-3 m+2 n=2\ \ldots\ (2)\)
\(p m-q n=5\ \ldots\ (3)\)
\(\text{Solve (1) and (2) by CAS:}\)
\(m=-\dfrac{8}{7}, n=-\dfrac{5}{7}\)
\(\text { Substitute } m,n \text { into (3): }\)
\begin{aligned}
-\dfrac{8}{7}\,p+\dfrac{5}{7}\,q & =5 \\
8 p & =5 q-35
\end{aligned}
\(\Rightarrow A\)
A ball is attached to the end of a string and rotated in a circle at a constant speed in a vertical plane, as shown in the diagram below.
The arrows in options \(\text{A}\). to \(\text{D}\). below indicate the direction and the size of the forces acting on the ball.
Ignoring air resistance, which one of the following best represents the forces acting on the ball when it is at the bottom of the circular path and moving to the left?
\(D\)
→ There are only two forces acting on the ball, gravitational force directly down, and the centripetal force (string tension) directly towards the centre of the circular motion.
→ As the ball is following a circular path, the tension in the string is greater than the gravitational force acting on the ball.
\(\Rightarrow D\)
A coil consisting of 20 loops with an area of 10 cm\(^2\) is placed in a uniform magnetic field \(B\) of strength 0.03 T so that the plane of the coil is perpendicular to the field direction, as shown in the diagram below.
The magnetic flux through the coil is closest to
\(B\)
→ Convert cm\(^2\) to m\(^2\): 1 cm\(^2\) = 0.01 m × 0.01 m = 0.0001 m\(^2\)
→ 10 cm\(^2\) = 0.0010 m\(^2\)
→ \(\Phi = BA = 0.03 \times 0.0010 = 3.0 \times 10^{-5}\ \text{Wb}\)
\(\Rightarrow B\)
A positron with a velocity of 1.4 × 10\(^6\) m s\(^{-1}\) is injected into a uniform magnetic field of 4.0 × 10\(^{-2}\) T, directed into the page, as shown in the diagram below.
It moves in a vacuum in a semicircle of radius \(r\). The mass of the positron is 9.1 × 10\(^{-31}\) kg and the charge on the positron is 1.6 × 10\(^{-19}\) C. Ignore relativistic effects.
Question 3
Which one of the following best gives the speed of the positron as it exits the magnetic field?
Question 4
The speed of the positron is changed to 7.0 × 10\(^5\) m s\(^{-1}\).
Which one of the following best gives the value of the radius \(r\) for this speed?
\(\text{Question 3:}\ C\)
\(\text{Question 4:}\ B\)
\(\text{Question 3}\)
→ As the direction of the force is perpendicular to the velocity, the velocity will not change.
→ The velocity will be \(1.4 \times 10^6\ \text{ms}^{-1}\)
\(\Rightarrow C\)
\(\text{Question 4}\)
\(F_B\) | \(=F_c\) | |
\(qvB\) | \(=\dfrac{mv^2}{r}\) | |
\(r\) | \(=\dfrac{mv}{qB}\) |
→ Therefore, \(r \propto v\).
→ If the velocity is halved, the radius will also be halved.
\(\Rightarrow B\)
Consider \(f:(-\infty, 1]\rightarrow R, f(x)=x^2-2x\). Part of the graph of \(y=f(x)\) is shown below.
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a. \([-1, \infty)\)
b.
c. \(\text{When }f(x)\ \text{is written in turning point form}\)
\(y=(x-1)^2-1\)
\(\text{Inverse function: swap}\ x \leftrightarrow y\)
\(x\) | \(=(y-1)^2-1\) |
\(x+1\) | \(=(y-1)^2\) |
\(-\sqrt{x+1}\) | \(=y-1\) |
\(f^{-1}(x)\) | \(=1-\sqrt{x+1}\) |
\(\text{Domain}\ [-1, \infty)\)
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a. \(\dfrac{1}{2}\)
b. \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)
a. | \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) | \(=\left[-\cos x\right]_0^\frac{\pi}{3}\) |
\(=-\cos\dfrac{\pi}{3}+\cos 0\) | ||
\(=-\dfrac{1}{2}+1\) | ||
\(=\dfrac{1}{2}\) |
b. | \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) | \(=\left[\sin x\right]_k^\frac{\pi}{2}\) |
\(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\) | ||
\(=1-\sin (k)\) |
\(\text{Using part (a):}\)
\(1-\sin (k)\) | \(=\dfrac{1}{2}\) |
\(\sin (k)\) | \(=\dfrac{1}{2}\) |
\(\therefore\ k\) | \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\) |
A photoelectric experiment is carried out by students. They measure the threshold frequency of light required for photoemission to be 6.5 × 10\(^{14}\) Hz and the work function to be 3.2 × 10\(^{-19}\) J.
Using the students' measurements, what value would they calculate for Planck's constant? Outline your reasoning and show all your working. Give your answer in joule-seconds. (3 marks)
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\(h=4.9 \times 10^{-34}\ \text{J s}\)
\(f=6.5 \times 10^{14}\ \text{Hz},\ \ E=3.2 \times 10^{-19}\ \text{J}\)
\(E=hf\ \ \Rightarrow\ \ h=\dfrac{E}{f}\)
\(\therefore h=\dfrac{3.2 \times 10^{-19}}{6.5 \times 10^{14}} = 4.92 \times 10^{-34}\ \text{J s}\)
On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.
The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.
One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.
The graph has been drawn to scale.
The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.
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The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.
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The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
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Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.
The population of rabbits over a longer period of time can be modelled by the rule
`s(t)=1700cdote^(-0.003t)cdot sin((pit)/80)+2500,\qquad text(for all)\ t>=0`
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ai. `r(0)=2500`
aii. Minimum population of rabbits `= 800`
Maximum population of rabbits `= 4200`
aiii. `160` weeks
b. See worked solution.
c. `~~ 5339` (nearest whole number)
d. Weeks between the periods is 160
e. `~~ 4142` (nearest whole number)
f. Average rate of change `=-3.6` rabbits/week (1 d.p.)
g. `t = 156` weeks (nearest whole number)
h. ` s → 2500`
ai. Initial population of rabbits
From graph when `t=0, \ r(0) = 2500`
Using formula when `t=0`
`r(t)` | `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500` | |
`r(0)` | `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500` rabbits |
aii. From graph,
Minimum population of rabbits `= 800`
Maximum population of rabbits `= 4200`
OR
Using formula
Minimum is when `t = 120`
`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`
Maximum is when `t = 40`
`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`
aiii. Number of weeks between maximum populations of rabbits `= 200-40 = 160` weeks
b. Period of foxes = period of rabbits = 160:
`frac{\2pi}{b} = 160`
`:.\ b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.
Using the point `(100 , 2500)`
Amplitude when `b = frac{\pi}{80}`:
`f(t)` | `=a \ sin (pi/80(t-60))+1600` | |
`f(100)` | `= 2500` | |
`2500` | `= a \ sin (pi/80(100-60))+1600` | |
`2500` | `= a \ sin (pi/2)+1600` | |
`a` | `= 2500-1600 = 900` |
`:.\ f(t)= 900 \ sin (pi/80)(t-60) + 1600`
c. Using CAS find `h(t) = f(t) + r(t)`:
`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`
`text{fMax}(h(t),t)|0 <= t <= 160` `t = 53.7306….`
`h(53.7306…)=5339.46`
Maximum combined population `~~ 5339` (nearest whole number)
d. Using CAS, check by changing domain to 0 to 320.
`text{fMax}(h(t),t)|0 <= t <= 320` `t = 213.7305…`
`h(213.7305…)=5339.4568….`
Therefore, the number of weeks between the periods is 160.
e. Fox population:
`t^{\prime} = frac{90}{pi}t + 60` → `t = frac{pi}{90}(t^{\prime}-60)`
`y^{\prime} = 900y+1600` → `y = frac{1}{900}(y^{\prime}-1600)`
`frac{y^{\prime}-1600}{900} = sin(frac{pi(t^{\prime}-60)}{90})`
`:.\ f(t) = 900\ sin\frac{pi}{90}(t-60) + 1600`
Average combined population [Using CAS]
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
`= 4142.2646….. ~~ 4142` (nearest whole number)
f. Using CAS
`s(t):= 1700e^(-0.003t) dot\sin\frac{pit}{80} + 2500`
`text{fMax}(s(t),t)|0<=t<=320` `x = 38.0584….`
`s(38.0584….)=4012.1666….`
`text{fMax}(s(t),t)|160<=t<=320` `x = 198.0584….`
`s(198.0584….)=3435.7035….`
Av rate of change between the points
`(38.058 , 4012.167)` and `(198.058 , 3435.704)`
`= frac{4012.1666….-3435.7035….}{38.0584….-198.0584….} =-3.60289….`
`:.` Average rate of change `=-3.6` rabbits/week (1 d.p.)
g. Using CAS
`s^(primeprime)(t) = 0` , `t = 80(n-0.049) \ \forall n \in Z`
After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2`
`t` | `= 80(n-0.049)` | |
`= 80(2-0.049)` | ||
`= 156.08` |
`:. \ t = 156` weeks (nearest whole number)
h. As `t → ∞`, `e^(-0.003t) → 0`
`:.\ s → 2500`
The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
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The tangent to `f` at point `M` has gradient `-2` .
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The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
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a. `x=0`
b. ` f^{\prime}(x)=1/6x`
c. `x=-12`
di. `y=1/2x + 18`
dii. Area`= 375` units²
e. `b = 2a^2`
a. Axis of symmetry: `x=0`
b. | `f(x)` | `=\frac{x^2}{12}` | |
` f^{\prime}(x)` | `= 1/6x` |
c. At `M` gradient `= -2`
`1/6x` | `= -2` | |
`x` | `= -12` |
When `x = -12`
`f(x) = (-12)^2/12 = 12`
Equation of tangent at `(-12 , 12)`:
`y-y_1` | `=m(x-x_1)` | |
`y-12` | `= -2(x + 12)` | |
`y` | `= -2x-12` |
d.i Gradient of tangent `= -2`
`:.` gradient of normal `= 1/2`
Equation at `M(- 12 , 12)`
`y -y_1` | `=m(x-x_1)` | |
`y-12` | `= 1/2(x + 12)` | |
`y` | `=1/2x + 18` |
d.ii Points of intersection of `f(x)` and normal are at `M` and `N`.
So equate ` y = x^2/12` and `y = 1/2x + 18` to find `N`
`x^2/12` | `=1/2x + 18` | |
`x^2-6x-216` | `=0` | |
`(x + 12)(x-18)` | `=0` |
`:.\ x = -12` or `x = 18`
Area | `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x` | |
`= [x^2/4 + 18x-x^3/36]_(-12)^18` | ||
`= [18^2/4 +18^2-18^3/36] – [12^2/4 + 18 xx (-12)-(-12)^3/36]` | ||
`= 375` units² |
e. `g(x) = x^2/(4a^2)` `a > 0`
At `x = -b` `y = (-b)^2/(4a^2) = b^2/4a^2`
`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`
Gradient of tangent `= (-b)/(2a^2)`
Gradient of normal `= (2a^2)/b`
Equation of normal at `(- b , b^2/(4a^2))`
`y-y_1` | `= m(x-x_1)` | |
`y-b^2/(4a^2)` | `= (2a^2)/b(x-(-b))` | |
`y` | `= (2a^2x)/b + 2a^2 + b^2/(4a^2)` | |
`y` | `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)` |
Points of intersection of normal and parabola (Using CAS)
solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`
`x =-b` or `x = (8a^4+b^2)/b`
Calculate area using CAS
`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2}-frac{x^2}{4a^2} \right) dx`
`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
Using CAS Solve derivative of `A = 0` with respect to `b` to find `b`
solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`
`b =-2a^2` and `b = 2a^2`
Given `b > 0`
`b = 2a^2`
The function `f(x)=\frac{1}{3} x^3+m x^2+n x+p`, for `m, n, p \in R`, has turning points at `x=-3` and `x=1` and passes through the point (3, 4).
The values of `m, n` and `p` respectively are
`B`
For turning point `f^{\prime}(x) = 0`
`f(x)` | `=\frac{1}{3} x^3 + mx^2 + nx + p` | |
`:.\ f^{\prime}(x)` | `= x^2-2mx+n` | |
` f^{\prime}(- 3)` | `= 9 – 6m + n` …. (1) | |
` f^{\prime}( 1)` | `= 1 + 2m + n` ….(2) |
Solving (1) and (2) | `9 – 6m + n` | `= 0` |
`1 + 2m + n` | `= 0` |
`m = 1` and `n = – 3`
`:. \ f(x)=\frac{1}{3} x^3 + x^2- 3x + p`
The point `(3 , 4)` lies on the line
`4` | `= \frac{1}{3} xx 3^3 +3^2- 3 xx 3 + p` | |
`:. \ p` | `= – 5 ` |
`:. \ f(x)=\frac{1}{3} x^3 + x^2- 3x – 5`
→ `m=1, n=-3, \quad p=-5`
`=>B`
A continuous random variable, \(X\), has a probability density function given by
\(f(x)=\begin{cases}\dfrac{2}{9}xe^{-\frac{1}{9}x^2} &\ \ x\geq 0 \\ \\ 0 &\ \ x<0 \\ \end{cases}\)
The expected value of \(X\), correct to three decimal places, is
\(B\)
\(\text{E}(X) = \displaystyle\int_0^{\infty} \dfrac{2}{9} x^2 e^{-\frac{1}{9} x^2}\,dx\) | \(\approx 2.65868\dots\) [by CAS] | |
\(\approx 2.659\) |
\(\Rightarrow B\)
If `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then `\frac{1}{k} \int x \ cos(x)dx` is equal to
`C`
Given `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then
`x\cdot \cos (x)` | `=\frac{d}{d x}(x\cdot \sin (x))-\sin (x)` | |
`\frac{1}{k} \int x \cos (x) d x` | `= \frac{1}{k}\left(\int \frac{d}{d x}(x \cdot \sin (x)) d x-\int \sin (x) d x\right)` | |
`= \frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x)d x\right)+c` |
`=> C`
The largest value of `a` such that the function `f:(-\infty, a] \rightarrow R, f(x)=x^2+3 x-10`, where `f` is one-to-one, is
`C`
`f(x)` | `= x^2+3 x-10` | |
`f^{\prime}(x)` | `= 2x + 3` |
Turning point when `f^{\prime}(x) = 0`
`\ 2x+3 = 0 \rightarrow x = -3/2`
`:.\ a = – 1.5`
`=>C`
Which one of the following functions is not continuous over the interval `x \in[0,5]`?
`D`
The gradient of the graph of `y=e^{3 x}` at the point where the graph crosses the vertical axis is equal to
`E`
Graph crosses vertical axis when `x = 0`
`y` | `= e^(3x)` |
`dy/dx` | `= 3e^(3x)` |
When `x = 0`, `m = 3 xx e^(3 xx 0) = 3`
`=>E`
In Young's double-slit experiment, the distance between two slits, S\(_1\) and S\(_2\), is 2.0 mm. The slits are 1.0 m from a screen on which an interference pattern is observed, as shown in Figure 13a. Figure 13b shows the central maximum of the observed interference pattern. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- a. \(3.1 \times 10^{-4}\ \text{m}\) b. The experiment supports the wave theory as follows: → Young’s double slit experiment demonstrates how light will refract and form an interference pattern against a screen. → Since interference is a wave phenomenon, the experiment supports the wave model of light. a. \(d\,\sin \theta=m \lambda\ \ \text{and}\ \ \sin \theta=\dfrac{\Delta x}{D}\) → \(\Delta x\) is the distance between two successive dark bands and \(D\) is the distance from the slits to the screen. → Young’s double slit experiment demonstrates how light will refract and form an interference pattern against a screen. → Since interference is a wave phenomenon, the experiment supports the wave model of light.
\(\dfrac{d\Delta x}{D}\)
\(=m \lambda\)
\(\Delta x\)
\(=\dfrac{D\lambda}{d}\)
\(=\dfrac{1 \times 620 \times 10^{-9}}{2 \times 10^{-3}}\)
\(=3.1 \times 10^{-4}\ \text{m}\)
b. The experiment supports the wave theory as follows:
Figure 11 shows a system of two ideal polarising filters, \(\text{F}_1\) and \(\text{F}_2\), in the path of an initially unpolarised light beam. The polarising axis of the first filter, \(\text{F}_1\), is parallel to the \(y\)-axis and the polarising axis of the second filter, \(\text{F}_2\), is parallel to the \(x\)-axis.
Will any light be observed at point \(\text{P}\)? Give your reasoning. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
→ Light will initially pass through \(F_1\) and so will be polarised in the vertical plane with the intensity of the light will reducing to 50%.
→ \(F_2\) will polarise light in the horizontal direction, but as the light beam is already polarised in the vertical direction, none of the light will be able to pass through \(F_2\).
→ Therefore, there will be no light visible at \(P\).
→ Light will initially pass through \(F_1\) and so will be polarised in the vertical plane with the intensity of the light will reducing to 50%.
→ \(F_2\) will polarise light in the horizontal direction, but as the light beam is already polarised in the vertical direction, none of the light will be able to pass through \(F_2\).
→ Therefore, there will be no light visible at \(P\).
A new spaceship that can travel at 0.7\(c\) has been constructed on Earth. A technician is observing the spaceship travelling past in space at 0.7\(c\), as shown in Figure 10. The technician notices that the length of the spaceship does not match the measurement taken when the spaceship was stationary in a laboratory, but its width matches the measurement taken in the laboratory. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. By Einstein’s theory of special relativity when travelling at high speeds, length contraction occurs. Length contraction only occurs in the direction of motion thus the technician measures a different measurement for the length but not the width of the spaceship. b. \(l_o=189\ \text{m}\) a. By Einstein’s theory of special relativity: → When travelling at high speeds, length contraction occurs. → Length contraction only occurs in the direction of motion thus the technician measures a different measurement for the length but not the width of the spaceship.
b.
\(l\)
\(=l_0\sqrt{1-\frac{v^2}{c^2}}\)
\(l_0\)
\(=\dfrac{l}{\sqrt{1-\frac{v^2}{c^2}}}\)
\(=\dfrac{135}{\sqrt{1-\frac{(0.7c)^2}{c^2}}}\)
\(=\dfrac{135}{\sqrt{1-0.7^2}}\)
\(=189\ \text{m}\)