A sequence of numbers is generated by the recurrence relation shown below.
\(R_0 = 2,\ \ \ R_{n+1} = 2-R_n\)
The value of \(R_2\) is
- \(-4\)
- \(-2\)
- \(0\)
- \(2\)
- \(4\)
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A sequence of numbers is generated by the recurrence relation shown below.
\(R_0 = 2,\ \ \ R_{n+1} = 2-R_n\)
The value of \(R_2\) is
\(D\)
\(R_0\) | \(=2\ \ \ \ \ R_{n+1} = 2-R_n\) | |
\(R_1\) | \(= 2-2 = 0\) | |
\(R_2\) | \(= 2-0 = 2\) |
\(\Rightarrow D\)
The seasonal index for sales of sunscreen in summer is 1.25
To correct for seasonality, the actual sunscreen sales for summer should be
\(A\)
\(\text{Deseasonalised Sales}\)
\(= \dfrac{\text{Actual Sales}}{\text{Seasonal Index}}\) | ||
\(=\dfrac{\text{Actual Sales}}{1.25}\) | ||
\(=0.8 \times \text{Actual Sales}\) |
\(\therefore \ \text{Summer sales should be reduced by 20%.}\)
\(\Rightarrow A\)
The scatterplot below displays the body length, in centimetres, of 17 crocodiles, plotted against their head length, in centimetres. A least squares line has been fitted to the scatterplot. The explanatory variable is head length.
Question 12
The equation of the least squares line is closest to
Question 13
The median head length of the 17 crocodiles, in centimetres, is closest to
Question 14
The correlation coefficient \(r\) is equal to 0.963
The percentage of variation in body length that is not explained by the variation in head length is closest to
\(\text{Question 12}:\ B\)
\(\text{Question 13}:\ B\)
\(\text{Question 12}:\ C\)
\(\text{Question 12}\)
\(\text{Gradient}\ = \dfrac{550-170}{85-30}=6.9\ \ \text{(eliminate D and E)}\)
\(\text{Head length is the explanatory variable (eliminate A and C)}\)
\(\Rightarrow B\)
\(\text{Question 13}\)
\(\text{Median score}\ =\dfrac{17+1}{2} = 9\text{th score}\)
\(\text{Median head length = 51 cm}\)
\(\Rightarrow B\)
\(\text{Question 14}\)
\(\text{Percentage explained by variation in head length}\)
\(r^2 = 0.963^2 = 0.9273 \approx 92.7\% \)
\(\text{Percentage not explained by variation in head length}\)
\(100-92.7 \approx 7.3\%\)
\(\Rightarrow C\)
Below is a reaction pathway beginning with hex-3-ene.
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a. 3-bromohexane
b. Steam and any specific inorganic strong acid (although not \(\ce{HCl}\)) is correct.
eg. \(\ce{H2O, H+}\)
c.
d. Correct answers included one of:
\(\ce{CH3COOH}\) or \(\ce{HOOCCH3}\)
e.i. Correct answers included one of the following:
\(\ce{CH3CH2COCH2CH2CH3}\)
\(\ce{CH3CH2CH2COCH2CH3}\)
\(\ce{CH3CH2CO(CH2)2CH3}\)
e.ii. Ketone
f. Oxidation
a. 3-bromohexane
b. Steam and any specific inorganic strong acid (although not \(\ce{HCl}\)) is correct.
eg. \(\ce{H2O, H+}\)
c.
d. Correct answers included one of:
\(\ce{CH3COOH}\) or \(\ce{HOOCCH3}\)
e.i. Correct answers included one of the following:
\(\ce{CH3CH2COCH2CH2CH3}\)
\(\ce{CH3CH2CH2COCH2CH3}\)
\(\ce{CH3CH2CO(CH2)2CH3}\)
e.ii. Ketone
f. Oxidation
How many structural isomers have the molecular formula \(\ce{C3H6BrCl}\)?
\(B\)
Isomers are:
\(\ce{CH3CH2CHBrCl}\): 1-bromo-1-chloropropane
\(\ce{CH3CHClCH2Br}\): 1-bromo-2-chloropropane
\(\ce{CH2ClCH2CH2Br}\): 1-bromo-3-chloroopropane
\(\ce{CH3CHBrCH2Cl}\): 2-bromo-1-chloropropane
\(\ce{CH3CBrClCH3}\): 2-bromo-2-chloropropane
\(\Rightarrow B\)
How does diluting a 0.1 M solution of lactic acid, \(\ce{HC3H5O3}\), change its pH and percentage ionisation?
pH | Percentage ionisation | |
A. | increase | decrease |
B. | increase | increase |
C. | decrease | increase |
D. | decrease | decrease |
\(B\)
→ Ionisation of lactic acid: \(\ce{HC3H5O3(aq) + H2O(l) \rightleftharpoons C3H5O3-(aq) + H3O+(aq)} \)
→ Adding water decreases \(\ce{[H3O+]}\). The equilibrium then shifts to partially compensate for this change, favouring the forward reaction.
→ \(\ce{[H3O+]}\) increases in this process although the new equilibrium \(\ce{[H3O+]}\) is lower (pH higher) than the original system.
→ Overall, the pH increases and the percentage ionisation increases.
\(\Rightarrow B\)
Butanoic acid is the simplest carboxylic acid that is also classified as a fatty acid. Butanoic acid may be synthesised as outlined in the following reaction flow chart. --- 0 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- i. ii. \(\ce{H2O}\) and \(\ce{H3PO4}\) (catalyst) iii. butan-1-ol or 1-butanol iv. \(\ce{CH3CH2CH2COOH}\) v. \(\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- \rightarrow 2Cr^{3+}(aq) + 7H2O(l)} \) i. iii. butan-1-ol or 1-butanol iv. \(\ce{CH3CH2CH2COOH}\) v. \(\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- \rightarrow 2Cr^{3+}(aq) + 7H2O(l)} \)
A student mixed salicylic acid with ethanoic anhydride (acetic anhydride) in the presence of concentrated sulfuric acid. The products of this reaction were the painkilling drug aspirin (acetyl salicylic acid) and ethanoic acid. --- 0 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- i. ii. Sulphuric acid increases the rate of reaction by: → providing an alternative reaction pathway that involves a lower activation energy for the reagents. → this increases the likelihood of successful collisions. i. ii. Sulphuric acid increases the rate of reaction by: → providing an alternative reaction pathway that involves a lower activation energy for the reagents. → this increases the likelihood of successful collisions.
Methanol is a liquid fuel that is often used in racing cars. The thermochemical equation for its complete combustion is
\(\ce{2CH3OH(l) + 3O2(g)\rightarrow 2CO2(g) + 4H2O(l) \quad \quad \ \ \ \Delta H = –1450 kJ mol^{–1}}\)
Octane is a principal constituent of petrol, which is used in many motor vehicles. The thermochemical equation for
the complete combustion of octane is
\(\ce{2C8H18(l) + 25O2(g)\rightarrow 16CO2(g) + 18H2O(l) \quad \quad \Delta H = –10\ 900 kJ mol^{-1}}\)
The molar mass of methanol is 32 g mol\(^{-1}\) and the molar mass of octane is 114 g mol\(^{–1}\). Which one of the following statements is the most correct?
\(B\)
Consider option B:
→ 1 mole of \(\ce{CH3OH}\) produces 725 kJ of heat energy
→ \(\ce{MM (CH3OH)} = 32.0\ \text{grams}\)
→ Heat energy of 1 gram \(\ce{CH3OH} = \dfrac{725}{32.0} = 22.7\) kJ
\(\Rightarrow B\)
When ethene is mixed with chlorine in the presence of UV light, the following reaction takes place.
\(\ce{CH2CH2(g) + Cl2(g) \xrightarrow{\text{UV light}} CH2ClCH2Cl(l)}\)
Reactions of organic compounds can be classified in a number of ways. The following list shows four possible classifications:
Which classification(s) applies to the reaction between ethene and chlorine?
\(C\)
Addition reaction:
→ \(\ce{Cl2}\) has added across the \(\ce{C=C}\) double bond (eliminate D).
Redox reaction:
→ The oxidation state of \(\ce{Cl}\) decreases from zero in \(\ce{Cl2(g)}\) to –1 \(\ce{CH2ClCH2Cl(l).}\)
→ In \(\ce{CH2=CH2}\), the oxidation state of each \(\ce{H}\) is +1, which means that the oxidation state of each \(\ce{C}\) is –2 .
→ In \(\ce{CH2ClCH2Cl}\), the oxidation states are \(\ce{H}:\) +1, \(\ce{Cl}:\) –1, \(\ce{C}:\) –1
→ The oxidation state of \(\ce{C}\) has increased (from –2 to –1 ) and the oxidation state of \(\ce{Cl}\) has decreased from 0 to –1 , confirming it is a redox reaction.
\(\Rightarrow C\)
What is the pH of a 0.0500 M solution of barium hydroxide, \(\ce{Ba(OH)2}\)?
\(D\)
\(\ce{Ba(OH)2(aq) \rightarrow Ba^{2+}(aq) + 2OH-(aq)} \)
\(\ce{[OH-] = 2 \times [Ba(OH)2] = 2 \times 0.0500 = 0.100 M} \)
\(\ce{[H3O+]} = \dfrac{10^{-14}}{\ce{[OH-]}} = \dfrac{10^{-14}}{0.100} = 10^{-13}\ \text{M} \)
\(\text{pH}\ = -\log_{10} 10^{-13} = 13\)
\(\Rightarrow D\)
Hydrogen sulfide, in solution, is a diprotic acid and ionises in two stages. \(\ce{H2S(aq) + H2O(l)\rightleftharpoons HS-(aq) + H3O+(aq)}\) \(\quad K_{a1} = 9.6 × 10^{–8} \text{ M}\) \(\ce{HS–(aq) + H2O(l)\rightleftharpoons S^{2-}(aq) + H3O+(aq)}\) \(\quad K_{a2} = 1.3 × 10^{–14} \text{ M}\) A student made two assumptions when estimating the pH of a \(0.01 \text{ M}\) solution of \(\ce{H2S}\): Assumption 1: The pH can be estimated by considering only the first ionisation reaction. Assumption 2: The concentration of \(\ce{H2S}\) at equilibrium is approximately equal to \(0.01 \text{ M}\). --- 6 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. 1st assumption: → \(K_{a2}\) is significantly smaller than the first ionisation \(\ce{(K_{a1})}\), making its impact on the \(\ce{[H3O+]}\) / pH level negligible. 2nd assumption: → \(K_{a1}\) is very small, making the extent of the ionisation of \(\ce{H2S}\) very small and hence a minimal change in \(\ce{[H2S]}\) results. b. \(\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 \) c. Adding \(\ce{NaHS}\): → Increases the \(\ce{[HS-]}\). → This increase causes the 1st ionisation equilibrium back to the left. → This left shift in the equilibrium decreases the \(\ce{[H3O+]}\) and the pH will therefore increase. a. 1st assumption: → \(K_{a2}\) is significantly smaller than the first ionisation \(\ce{(K_{a1})}\), making its impact on the \(\ce{[H3O+]}\) / pH level negligible. 2nd assumption: → \(K_{a1}\) is very small, making the extent of the ionisation of \(\ce{H2S}\) very small and hence a minimal change in \(\ce{[H2S]}\) results. \(=\dfrac{\ce{[H3O+]^2}}{0.01}\) c. Adding \(\ce{NaHS}\): → Increases the \(\ce{[HS-]}\). → This increase causes the 1st ionisation equilibrium back to the left. → This left shift in the equilibrium decreases the \(\ce{[H3O+]}\) and the pH will therefore increase.
b.
\(K_{a1}\)
\(=\dfrac{\ce{[HS-][H3O+]}}{\ce{[H2S]}} \)
\(9.6 \times 10^{-8}\)
\(\ce{[H3O+]^2}\)
\(=0.01 \times 9.6 \times 10^{-8} \)
\(\ce{[H3O+]}\)
\(=\sqrt{9.6 \times 10^{-10}}\)
\(=3.1 \times 10^{-5}\ \text{M} \)
\(\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 \)
The following table shows the value of the ionisation constant of pure water at various temperatures and at a constant pressure.
Temperature (°C) | 0 | 25 | 50 | 75 | 100 |
\(K_W\) | \( 1.1 \times 10^{-15}\) | \( 1.0 \times 10^{-14}\) | \( 5.5 \times 10^{-14}\) | \( 2.0 \times 10^{-13}\) | \( 5.6 \times 10^{-13}\) |
Given this data, which one of the following statements about pure water is correct?
\(B\)
\(\ce{2H2O(l) \rightleftharpoons H3O+(aq) + OH-(aq)} \)
→ Table shows that \(K_W\) increases as temperature increases.
→ Ionisation equation shifts right as temperature increases, causing an increase in \(\ce{[H3O+]}\) and \(\ce{[OH-]}\) (eliminate A).
→ pH decreases as \(\ce{[H3O+]}\) and temperature increase (eliminate C).
→ pH of water = 7 at 25°C only (eliminate D).
\(\Rightarrow B\)
A physics student comes across a river which runs north to south and has a current of 3 ms\(^{-1}\) running south.
The student starts on the west side of the river at point A and paddles a kayak at 5 ms\(^{-1}\) directly across the river to finish at point B.
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a. \(36.9^{\circ}\)
b. \(\text{25 seconds}\)
a.
\(\sin \theta\) | \(=\dfrac{3}{5}\) | |
\(\theta\) | \(=\sin^{-1}\Big(\dfrac{3}{5}\Big)=36.9^{\circ}\) |
The student must turn 36.9\(^{\circ}\) into the current as shown on the diagram.
b. Using Pythagoras:
\(v=\sqrt{5^2-3^2}=4\ \text{ms}^{-1}\)
\(\therefore\) It will take the student 25 seconds to travel from A to B.
Below is a description of the motion of a runner. The motion can be divided into three stages.
Stage 1: Runner travels 120 metres south taking 20 seconds.
Stage 2: Runner turns west and travels at 5 ms\(^{-1}\) for half a minute.
Stage 3: Runner travels directly back to their starting position.
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a. \(\text{150 m}\)
b. \(\text{192.1 m, N51.3°E.}\)
a. | \(d\) | \(=v \times t\) |
\(=5 \times 30\) | ||
\(=150\ \text{m}\) |
b. Stage 1 and Stage 2 displacement diagram:
Using Pythagoras:
\(d\) | \(=\sqrt{150^2 + 120^2} \) | |
\(=\sqrt{36900}\) | ||
\(=192.1\ \text{m}\) |
\(\tan\,\theta\) | \(=\dfrac{120}{150} \) | |
\(\theta\) | \(=\tan^{-1}\Big(\dfrac{120}{150}\Big)\) | |
\(=38.7^{\circ}\) |
\(\therefore\) Displacement of the start point from the runner is 192.1 m, N51.3°E.
The histogram below displays the distribution of skull width, in millimetres, for 46 female possums.
Question 1
The shape of the distribution is best described as
Question 2
The percentage of the 46 possums with a skull width of less than 55 mm is closest to
Question 3
The third quartile \((Q_3)\) for this distribution, in millimetres, could be
\(\text{Question 1:} \ C\)
\(\text{Question 2:} \ B\)
\(\text{Question 3:} \ D\)
\(\text{Question 1}\)
The distribution’s centre is in the 56–57 group and if the possible outlier is disregarded, the tail of the distribution is spread more to the left → i.e. negatively skewed with possible outlier.
\(\Rightarrow C\)
\(\text{Question 2}\)
\(\text{Percentage}\) | \(= \dfrac{1+1+2+1+2+5}{46} \times 100\) | |
\(=\dfrac{12}{46} \times 100\) | ||
\(=26.086…\)% |
\(\Rightarrow B\)
\(\text{Question 3}\)
\( Q_3\ =0.75 \times 46 = 34.5 \)
The 34.5th score lies between 57 and 58, therefore, 57.7.
\(\Rightarrow D\)
A plane is travelling at 315 ms\(^{-1}\) north when it passes through a dense cloud and slows down to a velocity of 265 ms\(^{-1}\) for safety precautions.
The plane did not change direction and travelled 2.5 km while it was slowing down.
Using north as the positive direction for all calculations, determine:
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a. \(\text{50 ms}^{-1}\ \text{south}\)
b. \(\text{5.8 ms}^{-2}\ \text{south}\)
c. \(\text{8.62 s}\)
a. | \(\Delta v\) | \(=v-u\) |
\(=265-315\) | ||
\(=-50\ \text{ms}^{-1}\) | ||
\(=50\ \text{ms}^{-1}\ \text{south}\) |
b. Using \(v^2=u^2 +2as\) (time is not given):
\(a\) | \(=\dfrac{v^2-u^2}{2s}\) | |
\(=\dfrac{(265)^2-(315)^2}{2 \times 2500}\) | ||
\(=-5.8\ \text{ms}^{-2}\) | ||
\(=5.8\ \text{ms}^{-2}\) to the south. |
c. Using \(v=u+at\):
\(t\) | \(=\dfrac{v-u}{a}\) | |
\(=\dfrac{265-315}{-5.8}\) | ||
\(=8.62\ \text{s}\) |
Plane A is flying due north at 300 kmh\(^{-1}\) when it measures the velocity of plane B flying due south to be 750 kmh\(^{-1}\).
Calculate the velocity of plane B as measured by the pilots on plane B? (3 marks)
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\(\text{ 450 kmh}^{-1}\ \text{south.} \)
Let north be the positive direction and south be the negative direction.
\(v_{\text{B rel A}}\) | \(=v_{\text{B}}-\ v_{\text{A}}\) | |
\(-750\) | \(=v_B-300\) | |
\(v_B\) | \(=-450\) | |
\(=450\ \text{kmh}^{-1}\ \text{south} \) |
A hot-air balloon is travelling at a constant upwards velocity of 15 ms\(^{-1}\).
A passenger on the hot-air balloon decides to time how long it takes a pen to hit the ground when dropped from a height of 50 m.
Ignoring air resistance, determine how long it will take the pen to hit the ground. (4 marks)
\(5.07\ \text{s}\)
\(s= 50, \ a=9.8, \ u=-15\)
Downwards \(\Rightarrow\) positive direction
\(s\) | \(=ut+\dfrac{1}{2}at^2\) | |
\(50\) | \(=-15t + \dfrac{1}{2} \times 9.8 \times t^2\) | |
\(0\) | \(=4.9t^2-15t-50\) | |
\(t\) | \(=\dfrac{15\pm \sqrt{225-4 \times 9.8 \times -50}}{2 \times 4.9}\) | |
\(=5.07, -2.01\) |
Time for the pen to fall = 5.07 seconds \( (t \gt 0) \).
The network below shows the one-way paths between the entrance, \(A\), and the exit, \(H\), of a children's maze.
The vertices represent the intersections of the one-way paths.
The number on each edge is the maximum number of children who are allowed to travel along that path per minute.
The minimum cut of the network is drawn, showing the maximum flow capacity of the maze is 23 children per minute.
One path in the maze is to be changed.
Determine the changes in the maximum flow capacity of the network in each of the following changes
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i. \(GH ↑ 16,\ \text{minimum cut = 27}\)
\(\text{Change: increases by 4}\)
ii. \(CE ↑ 12,\ \text{minimum cut = 24}\)
\(\text{Change: increases by 1}\)
iii. \(GF\ \text{is reversed, minimum cut = 30 (close to exit H)}\)
\(\text{Change: increases by 7}\)
i. \(GH ↑ 16,\ \text{minimum cut = 27}\)
\(\text{Change: increases by 4}\)
ii. \(CE ↑ 12,\ \text{minimum cut = 24}\)
\(\text{Change: increases by 1}\)
iii. \(GF\ \text{is reversed, minimum cut = 30 (close to exit H)}\)
\(\text{Change: increases by 7}\)
The network below shows the one-way paths between the entrance, \(A\), and the exit, \(H\), of a children's maze.
The vertices represent the intersections of the one-way paths.
The number on each edge is the maximum number of children who are allowed to travel along that path per minute.
Cuts on this network are used to consider the possible flow of children through the maze.
Determine the capacity of the minimum cut of this network. (2 marks)
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\(\text{Minimum cut = 23} \)
\(\text{Minimum cut}\ = 12+4+7 = 23\)
The network below shows the one-way paths between the entrance, \(A\), and the exit, \(H\), of a children's maze.
The vertices represent the intersections of the one-way paths.
The number on each edge is the maximum number of children who are allowed to travel along that path per minute.
Question 39
Cuts on this network are used to consider the possible flow of children through the maze. The capacity of the minimum cut would be
Question 40
One path in the maze is to be changed.
Which one of these five changes would lead to the largest increase in flow from entrance to exit?
\(\text{Question 39:}\ B \)
\(\text{Question 40:}\ E \)
\(\text{Question 39} \)
\(\text{Minimum cut}\ = 12+4+7 = 23\)
\(\Rightarrow B\)
\(\text{Question 40}\)
\(CE ↑ 12,\ \text{minimum cut = 24}\)
\(FH ↑ 14,\ \text{minimum cut = 23}\)
\(GH ↑ 16,\ \text{minimum cut = 27}\)
\(CF\ \text{is reversed, minimum cut = 29}\)
\(GF\ \text{is reversed, minimum cut = 30 (close to exit H)}\)
\(\Rightarrow E\)
The adjacency matrix below represents a planar graph with five vertices.
\begin{aligned}
& \ \ \ J\ \ \ K\ \ \ L\ \ M\ \ N \\
& {\left[\begin{array}{lllll}
0 & 1 & 0 & 1 & 1 \\
1 & 0 & 2 & 1 & 1 \\
0 & 2 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 0
\end{array}\right] \begin{array}{l}
J \\
K \\
L \\
M \\
N
\end{array}} \\
\end{aligned}
The number of faces on the planar graph is
\(B\)
\(\text{Sketch network:}\)
\(\Rightarrow B\)
Four employees, Anthea, Bob, Cho and Dario, are each assigned a different duty by their manager.
The time taken for each employee to complete duties 1,2,3 and 4, in minutes, is shown in the table below
\begin{array} {|l|c|c|c|c|}
\hline \rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{Duty 1} & \text{Duty 2} & \text{Duty 3} & \text{Duty 4} \\
\hline \rule{0pt}{2.5ex} \text{Anthea} \rule[-1ex]{0pt}{0pt} & \text{8} & \text{7} & \text{7} & \text{8}\\
\hline \rule{0pt}{2.5ex} \text{Bob} \rule[-1ex]{0pt}{0pt} & \text{10} & \text{8} & \text{10} & \text{9}\\
\hline \rule{0pt}{2.5ex} \text{Cho} \rule[-1ex]{0pt}{0pt} & \text{8} & \text{9} & \text{7} & \text{10}\\
\hline \rule{0pt}{2.5ex} \text{Dario} \rule[-1ex]{0pt}{0pt} & \text{7} & \text{7} & \text{8} & \text{9}\\
\hline
\end{array}
The manager allocates the duties so as to minimise the total time taken to complete the four duties.
The minimum total time taken to complete the four duties, in minutes, is
\(B\)
\(\text{By CAS:}\)
\begin{array} {|l|c|c|c|c|}
\hline \rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{Duty 1} & \text{Duty 2} & \text{Duty 3} & \text{Duty 4} \\
\hline \rule{0pt}{2.5ex} \text{Anthea} \rule[-1ex]{0pt}{0pt} & \text{8} & \textbf{[7]} & \text{7} & \text{8}\\
\hline \rule{0pt}{2.5ex} \text{Bob} \rule[-1ex]{0pt}{0pt} & \text{10} & \text{8} & \text{10} & \textbf{[9]}\\
\hline \rule{0pt}{2.5ex} \text{Cho} \rule[-1ex]{0pt}{0pt} & \text{8} & \text{9} & \textbf{[7]} & \text{10}\\
\hline \rule{0pt}{2.5ex} \text{Dario} \rule[-1ex]{0pt}{0pt} & \text{7} & \textbf{[7]} & \text{8} & \text{9}\\
\hline
\end{array}
\(\text{Minimum time}\ = 7+9+7+7 = 30\ \text{mins} \)
\(\Rightarrow B\)
For one particular week in a school year, students at Phyllis Island Primary School can spend their lunch break at the playground \((P)\), basketball courts \((B)\), oval \((O)\) or the library \((L)\).
Students stay at the same location for the entire lunch break.
The transition diagram below shows the proportion of students who change location from one day to the next.
The transition diagram is incomplete.
On the Monday, 150 students spent their lunch break at the playground, 50 students spent it at the basketball courts, 220 students spent it at the oval, and 40 students spent it in the library.
Of the students expected to spend their lunch break on the oval on the Wednesday, the percentage of these students who also spent their lunch break on the oval on Tuesday is closest to
\(C\)
\(\text{Transition matrix}\ (T): \)
\begin{aligned}
& \quad \quad \quad \ \ \ \ \textit{today} \\
& \quad \ \ \ \ P \ \ \ \ \ B \ \ \ \ \ \ O \ \ \ \ \ \ L \ \ \\
\ \ \textit{next day}\ \ \ & \begin{array}{l}
P \\
B \\
O \\
L
\end{array}\begin{bmatrix}
0.2 & 0.4 & 0.3 & 0.4 \\
0.3 & 0.3 & 0.3 & 0.2 \\
0.4 & 0.2 & 0.3 & 0.1 \\
0.1 & 0.1 & 0.1 & 0.3
\end{bmatrix}
\begin{bmatrix}
150 \\
50 \\
220 \\
40
\end{bmatrix}
= \begin{bmatrix}
132 \\
134 \\
140 \\
54 \\
\end{bmatrix}
\end{aligned}
\(T \times \begin{bmatrix}
132 \\
134 \\
140 \\
54
\end{bmatrix}
= \begin{bmatrix}
143.6 \\
132.6 \\
127 \\
56.8
\end{bmatrix}\)
\(\text{Students at oval Tue and Wed}\ = 0.3 \times 140 = 42\)
\(\text{Percentage}\ = \dfrac{42}{127} = 0.3307 \approx 33\% \)
\(\Rightarrow C\)
A species of bird has a life span of three years.
The females in this species do not reproduce in their first year but produce an average of four female offspring in their second year, and three in their third year.
The Leslie matrix, \(L\), below is used to model the female population distribution of this species of bird.
\(L=\begin{bmatrix}
0 & 4 & 3\\
0.2 & 0 & 0\\
0 & 0.4 & 0
\end{bmatrix}\)
The element in the second row, first column states that on average 20% of this population will
\(C\)
\(\text{Second and third rows represent survival rates from one year to the next.} \)
\(e_{21}\ \text{indicates an average of 20% survive into their second year.}\)
\(\Rightarrow C\)
How many of the following statements are true?
\(D\)
\(\text{All square matrices have an inverse → not true}\)
\(\text{The inverse of a matrix could be the same as the transpose of that matrix → true}\)
\(\text{If the determinant of a matrix is equal to zero, then the inverse does not exist → true}\)
\(\text{It is possible to take the inverse of an identity matrix → true}\)
\(\Rightarrow D\)
--- 8 WORK AREA LINES (style=lined) ---
\( e^{n^2+n}>(n !)^2 .\) (3 marks)
--- 11 WORK AREA LINES (style=lined) ---
i. \(\text{See Worked Solutions}\)
ii. \(\text{See Worked Solutions}\)
i. \(\text{Prove}\ \ x > \ln x\ \ \text{for} \ \ x>0: \)
\(\Rightarrow \ \text{Show}\ \ f(x) = x-\ln x > 0 \)
\(\text{SP’s occur when}\ \ f^{′}(x) = 1-\dfrac{1}{x} = 0\)
\(\text{SP at}\ (1,1) \)
\(f^{″} = x^{-2}>0,\ \ \forall x>0 \)
\(\text{SP at (1, 1) is a global minimum for}\ x>0 \)
\(\Rightarrow f(x) \geq 1 > 0 \)
\(\therefore x > \ln x\ \ \text{for} \ \ x>0 \)
ii. \(x > \ln x\ \ \text{for} \ \ x>0 \ \ \Rightarrow \ \ e^x > x\ \ \text{(by definition)} \)
\(\text{Choose any positive integer}\ n: \)
\(e^n\) | \(>n \) | |
\(e^{n-1}\) | \(>n-1 \) | |
\(\ \ \vdots \) | ||
\(e^2\) | \(>2\) | |
\(e^1\) | \(>1\) |
\(\text{Multiply each side of the equations above:}\)
\(e^n \times e^{n-1} \times \cdots \times e^{1} \) | \(>n(n-1)(n-2) \cdots (2)(1) \) | |
\(e^{n+(n-1)+(n-2)+ \cdots + 2+1}\) | \(>n!\) | |
\(e^{\frac{n(n+1)}{2}} \) | \(>n!\ \ (\text{using AP formula}\ \ S_n=\frac{n}{2}(a+l) ) \) | |
\(e^{\frac{n(n+1)}{2} \times 2}\) | \(>(n!)^2\) | |
\(e^{n^2+n}\) | \(>(n!)^2\ \ …\ \text{as required}\) |
Complete the table below, describing the reactivity characteristics of the three metals listed. (3 marks)
\begin{array} {|l|l|l|l|}
\hline
\ \ \ \ \textbf{Metal} \ & \ \ \ \ \textbf{Reactivity with} \ \ \ \ &\ \ \ \ \textbf{Reactivity with}\ \ \ \ & \ \ \ \ \textbf{Reactivity with}\ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \textbf{water} & \ \ \ \ \ \ \ \ \textbf{dilute acid} & \ \ \ \ \ \ \ \ \textbf{oxygen}\\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & & & \\ \rule{0pt}{2.5ex} \text{Potassium (K)} \rule[-1ex]{0pt}{0pt} & & & \\ & & & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & & & \\ \rule{0pt}{2.5ex} \text{Zinc (Zn)} \rule[-1ex]{0pt}{0pt} & & & \\ & & & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & & & \\ \rule{0pt}{2.5ex} \text{Copper (Cu)} \rule[-1ex]{0pt}{0pt} & & & \\ & & & \\
\hline
\end{array}
--- 0 WORK AREA LINES (style=lined) ---
\begin{array} {|l|l|l|l|}
\hline
\ \ \ \ \textbf{Metal} \ & \ \ \ \ \textbf{Reactivity with} \ \ \ \ &\ \ \ \ \textbf{Reactivity with}\ \ \ \ & \ \ \ \ \textbf{Reactivity with}\ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \textbf{water} & \ \ \ \ \ \ \ \ \textbf{dilute acid} & \ \ \ \ \ \ \ \ \textbf{oxygen}\\
\hline
& \text{Violent reaction that} & \text{Highly exothermic,} & \text{Burns rapidly to form} \\ \text{Potassium (K)} & \text{ignites hydrogen gas} & \text{ignition of the hydrogen} & \text{oxide which combusts}\\ & \text{produced.} & \text{produced.} & \text{spontaneously in air.} \\
\hline
& \text{Reacts slower with no} & \text{Bubbles slowly to} & \text{Burns when heated with} \\ \text{Zinc (Zn)} & \text{combustion (can be sped} & \text{moderately with no} & \text{oxygen, forming a less} \\ & \text{up using steam).} & \text{ignition.} & \text{reactive oxide layer (may}\\ & & & \text{further react with water).} \\
\hline
& & & \text{Slowly reacts, producing} \\ \text{Copper (Cu)} & \text{No reaction.} & \text{No reaction.} & \text{highly stable oxide layer} \\ & & & \text{that prevents further} \\ & & & \text{oxidation.} \\
\hline
\end{array}
\begin{array} {|l|l|l|l|}
\hline
\ \ \ \ \textbf{Metal} \ & \ \ \ \ \textbf{Reactivity with} \ \ \ \ &\ \ \ \ \textbf{Reactivity with}\ \ \ \ & \ \ \ \ \textbf{Reactivity with}\ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \textbf{water} & \ \ \ \ \ \ \ \ \textbf{dilute acid} & \ \ \ \ \ \ \ \ \textbf{oxygen}\\
\hline
& \text{Violent reaction that} & \text{Highly exothermic,} & \text{Burns rapidly to form} \\ \text{Potassium (K)} & \text{ignites hydrogen gas} & \text{ignition of the hydrogen} & \text{oxide which combusts}\\ & \text{produced.} & \text{produced.} & \text{spontaneously in air.} \\
\hline
& \text{Reacts slower with no} & \text{Bubbles slowly to} & \text{Burns when heated with} \\ \text{Zinc (Zn)} & \text{combustion (can be sped} & \text{moderately with no} & \text{oxygen, forming a less} \\ & \text{up using steam).} & \text{ignition.} & \text{reactive oxide layer (may}\\ & & & \text{further react with water).} \\
\hline
& & & \text{Slowly reacts, producing} \\ \text{Copper (Cu)} & \text{No reaction.} & \text{No reaction.} & \text{highly stable oxide layer} \\ & & & \text{that prevents further} \\ & & & \text{oxidation.} \\
\hline
\end{array}
Metals such as Lead, Copper, Mercury and Silver do not react with dilute acids but will react with the same acids at higher concentration levels.
Explain why this occurs with reference to first ionisation energy. (3 marks)
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→ First ionsiation energy refers to the energy required to remove a valence electron from a metal.
→ Metals react in their ionised state and the less energy that is required to reach this state, the more reactive the metal.
→ Lead, Copper, Mercury and Silver require more energy to remove their outer valence electrons and consequently are only reactive in the presence of stronger oxidising agents.
→ Concentrated acids have more oxidising ability than their dilute counterparts and can react with these metals.
→ First ionsiation energy refers to the energy required to remove a valence electron from a metal.
→ Metals react in their ionised state and the less energy that is required to reach this state, the more reactive the metal.
→ Lead, Copper, Mercury and Silver require more energy to remove their outer valence electrons and consequently are only reactive in the presence of stronger oxidising agents.
→ Concentrated acids have more oxidising ability than their dilute counterparts and can react with these metals.
In a laboratory, students reacted aluminium with water to produce an oxide and hydrogen gas.
Which of the following equations correctly represents this reaction.
\(D\)
→ Aluminium is not reactive enough to form an oxide with liquid water over a short period. For this to occur additional energy needs to be input (eliminate B).
→ This can however be achieved by reacting the aluminium with steam, which has more energy than water.
→ Option D provides the only balanced equation of the remaining options.
\(\Rightarrow D\)
Two moles of butane \(\ce{C3H8(g)}\) were reacted with 224 grams of oxygen \(\ce{O2(g)}\).
--- 6 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i. \(\ce{2C3H8(g) + 7O2(g)\ \rightarrow 2C(s) + 2CO(g) + 2CO2(g) + 8H2O(l)}\)
ii. \(\ce{m(CO2) = 88.02\ \text{g}}\)
i. Complete combustion equation:
\(\ce{C3H8(g) + 5O2(g)\ \rightarrow 3CO2(g) + 4H2O(l)} \)
→ Two moles of butane require 10 moles of oxygen to fully combust.
→ \(\ce{n(O2) = \dfrac {m}{MM}= \dfrac {224}{32} = 7}\)
→ Oxygen is limiting reagent and butane will undergo incomplete combustion according to the following balanced equation:
\(\ce{2C3H8(g) + 7O2(g)\ \rightarrow 2C(s) + 2CO(g) + 2CO2(g) + 8H2O(l)}\)
ii. Using the equation in part (i), 2 moles of \(\ce{CO2}\) will be produced
\(\ce{m(CO2) = n \times MM = 2 \times 44.01 = 88.02\ \text{g}}\)
The following recurrence relation models the value, \(P_n\), of a perpetuity after \(n\) time periods.
\(P_0=a, \quad P_{n+1}=R P_n-d\)
The value of \(R\) can be found by calculating
\(B\)
\(P_0=a, \quad P_{n+1}=R P_n-d\)
\(P_1=R P_0-d\)
\(\text{Since}\ \ P_0=P_1\ \ \text{(perpetuities retain same value)} \)
\(a\) | \(=Ra-d\) | |
\(Ra\) | \(=a+d\) | |
\(R\) | \(=\dfrac{a+d}{a} \) |
\(\Rightarrow B\)
Timmy took out a reducing balance loan of $500 000, with interest calculated monthly.
The balance of the loan, in dollars, after \(n\) months, \(T_n\), can be modelled by the recurrence relation
\(T_0=500\ 000, \quad T_{n+1}=1.00325 T_n-2611.65\)
A final repayment that will fully repay the loan to the nearest cent is
\(D\)
\(r \text{(annual)}\ = 12 \times 0.00325 = 3.9\% \)
\(\text{Find}\ N \ \text{when}\ FV=0\ \text{(by TVM solver)}:\)
\(N\) | \(= ?\) |
\(I(\%)\) | \(= 3.9\) |
\(PV\) | \(= -500\ 000\) |
\(PMT\) | \(= 2611.65\) |
\(FV\) | \(= 0\) |
\(\text{P/Y}\) | \(=\ \text{C/Y}\ = 12\) |
\(N=300.00094…\)
\(\text{Find}\ FV \ \text{when}\ N=300\ \text{(by TVM solver)}:\)
\(N\) | \(= 300\) |
\(I(\%)\) | \(= 3.9\) |
\(PV\) | \(= -500\ 000\) |
\(PMT\) | \(= 2611.65\) |
\(FV\) | \(= ?\) |
\(\text{P/Y}\) | \(=\ \text{C/Y}\ = 12\) |
\(FV=-2.466\)
\(\text{Final payment}\ = 2611.65+2.47 = $2614.12\)
\(\Rightarrow D\)
For taxation purposes, Audrey depreciates the value of her $3000 computer over a four-year period. At the end of the four years, the value of the computer is $600.
Question 20
If Audrey uses flat rate depreciation, the depreciation rate, per annum is
Question 21
If Audrey uses reducing balance depreciation, the depreciation rate, per annum is closest to
\(\text{Question 20:}\ C\)
\(\text{Question 21:}\ E\)
\(\text{Question 20}\)
\(\text{Depreciation value}\ = 3000-600=$2400 \)
\(\text{Depreciation value (per year)}\ = \dfrac{2400}{4} =$600 \)
\(\text{Depreciation rate}\ = \dfrac{600}{3000} \times 100 =20\% \)
\(\Rightarrow C\)
\(\text{Question 21}\)
\(A = $600, \ P= $3000,\ n=4\)
\(A\) | \(=PR^n\) | |
\(600\) | \(=3000 \times R^4\) | |
\(R^4\) | \(=\dfrac{600}{3000} \) | |
\(R\) | \(=\sqrt[4]{0.2} \) | |
\(=0.668…\) |
\(\text{Depreciation rate}\ =1-0.668… = 0.331… \approx 33\% \)
\(\Rightarrow E\)
The number of visitors each month to a zoo is seasonal.
To correct the number of visitors in January for seasonality, the actual number of visitors, to the nearest percent, is increased by 35%.
The seasonal index for that month is closest to
\(D\)
\(\text{Seasonal index}\) | \(=\ \dfrac{\text{actual}}{\text{deseasonalised}} \) | |
\(=\ \dfrac{\text{actual}}{1.35 \times \text{actual}} \) | ||
\(=0.741\) |
\(\Rightarrow D\)
The number of visitors to a public library each day for 10 consecutive days was recorded.
These results are shown in the table below.
\begin{array} {|l|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Day number} \rule[-1ex]{0pt}{0pt} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex} \textbf{Number of visitors} \rule[-1ex]{0pt}{0pt} & 337 & 317 & 313 & 335 & 322 & 335 & 322 & 338 & 302 & 349 \\
\hline
\end{array}
The eight-mean smoothed number of visitors with centring for day number 6 is
\(C\)
\(\text{8 Mean average (days 2 – 9):}\)
\((317+313+335+322+335+322+338+302)\ ÷\ 8 = 323 \)
\(\text{8 Mean average (days 3 – 10):}\)
\((313+335+322+335+322+338+302+349)\ ÷\ 8 = 327 \)
\(\text{Centred value}\ = \dfrac{323+327}{2} = 325\)
\(\Rightarrow C\)
The following graph shows a selection of winning times, in seconds, for the women's 800 m track event from various athletic events worldwide. The graph shows one winning time for each calendar year from 2000 to 2022.
Question 13
The time series is smoothed using seven-median smoothing.
The smoothed value for the winning time in 2006, in seconds, is closest to
Question 14
The median winning time, in seconds, for all the calendar years from 2000 to 2022 is closest to
\(\text{Question 13:}\ C\)
\(\text{Question 14:}\ B\)
\(\text{Question 13}\)
\(\text{Consider the 2006 data point and 3 data points either side.}\)
\(\text{Median value (of 7 data points) = 116.8}\)
\(\Rightarrow C\)
\(\text{Question 14}\)
\(\text{23 data points between 2000 – 2022.}\)
\(\text{Median value = 12th data point (in order) = 117.2}\)
\(\Rightarrow B\)
A least squares line can be used to model the birth rate (children per 1000 population) in a country from the average daily food energy intake (megajoules) in that country.
When a least squares line is fitted to data from a selection of countries it is found that:
The slope of this least squares line is closest to
\(B\)
\(\text{Independent variable}\ (x)\ \text{= energy intake} \)
\(\text{Dependent variable}\ (y)\ \text{= birth rate} \)
\(\text{LSLR passes through (8.53, 32.2) and (14.9, 9.9) }\)
\(\text{Slope of LSRL}\ = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{32.2-9.9}{8.53-14.9} = -3.501 \)
\(\Rightarrow B\)
A teacher analysed the class marks of 15 students who sat two tests.
The test 1 mark and test 2 mark, all whole number values, are shown in the scatterplot below.
A least squares line has been fitted to the scatterplot.
Question 7
The equation of the least squares line is closest to
Question 8
The least squares line shows the predicted test 2 mark for each student based on their test 1 mark.
The number of students whose actual test 2 mark was within two marks of that predicted by the line is
\(\text{Question 7:}\ A\)
\(\text{Question 8:}\ C\)
\(\text{Question 7}\)
\(\text{By inspection, gradient is greater than 1 (eliminate B and C)}\)
\(\text{LSRL passes through (16, 18):}\)
\(\text{Option A:}\ -6.83 + 1.55 \times 16 = 18.0\ \checkmark \)
\(\Rightarrow A\)
\(\text{Question 8}\)
\(\text{5 values are within 1 grid height (measured vertically), or 2 marks,}\)
\(\text{from the LRSR.}\)
\(\Rightarrow C\)
The heights of a group of Year 8 students have a mean of $163.56 cm and a standard deviation of $8.14 cm.
One student's height has a standardised \(z\)-score of –0.85 .
This student's height, in centimetres, is closest to
\(B\)
\(\bar x = 163.56, \ s_x = 8.14\)
\(\text{Find}\ x\ \text{given}\ \ z=-0.85:\)
\(z\) | \(= \dfrac{x-\bar x}{s_x}\) | |
\(-0.85\) | \(=\dfrac{x-163.56}{8.14}\) | |
\(x-163.56\) | \(=8.14 \times -0.85\) | |
\(x\) | \(=163.56-6.919\) | |
\(=156.641\) |
\(\Rightarrow B\)
In the system diagram below, a 5-kilogram mass and masses \(A\) and \(B\) are held by high tensile frictionless wire in static equilibrium.
Using a vector diagram, calculate the masses of both \(A\) and \(B\). (4 mark)
--- 8 WORK AREA LINES (style=blank) ---
\(\text{Mass}_A =3.91\ \text{kg}\)
\(\text{Mass}_B =4.55\ \text{kg}\)
Using the sin rule both \(F_B\) and \(F_A\) can be calculated:
\(\dfrac{F_A}{\sin 48^{\circ}}\) | \(=\dfrac{5 \times 9.8}{\sin 72^{\circ}}\) | |
\(F_A\) | \(=\dfrac{49\,\sin 48^{\circ}}{\sin 72^{\circ}}\) | |
\(=38.3\ \text{N}\) |
\(\text{Mass}_A =\dfrac{F}{a}=\dfrac{38.3}{9.8}=3.91\ \text{kg}\)
\(\dfrac{F_B}{\sin 60^{\circ}}\) | \(=\dfrac{49}{\sin 72^{\circ}}\) |
\(F_B\) | \(=\dfrac{49\, \sin 60^{\circ}}{\sin 72^{\circ}}\) |
\(=44.6\ \text{N}\) |
\(\text{Mass}_B =\dfrac{F}{a}=\dfrac{44.6}{9.8}=4.55\ \text{kg}\)
Outline and explain the process by which ferromagnetic materials can become magnetised using a bar magnet. (3 marks)
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→ A ferromagnetic material contains many free electrons due to its metallic bonding.
→ Each electron has its own magnetic domain but due to the electrons spinning in opposite directions, the domains cancel each other out.
→ Using a bar magnet, stroke the ferromagnetic material in one direction.
→ This causes the electrons to spin in the same direction, hence all of their domains line up.
→ The cumulative effect of all the small magnetic domains creates a large magnetic field, turning the ferromagnetic material into a magnet.
→ A ferromagnetic material contains many free electrons due to its metallic bonding.
→ Each electron has its own magnetic domain but due to the electrons spinning in opposite directions, the domains cancel each other out.
→ Using a bar magnet, stroke the ferromagnetic material in one direction.
→ This causes the electrons to spin in the same direction, hence all of their domains line up.
→ The cumulative effect of all the small magnetic domains creates a large magnetic field, turning the ferromagnetic material into a magnet.
A wire with a current of \(I\) amps running through it was measured to have a magnetic field strength of 2 × 10\(^{-3}\) T at a distance of \( r\) metres from the wire.
If the current through the wire is halved and the distance \(r\) is increased to \(3r\), find the new magnetic field strength measured. (2 marks)
--- 4 WORK AREA LINES (style=blank) ---
\(3.33 \times 10^{-4}\ \text{T}\)
\(B_{\text{initial}}= 2 \times 10^{-3} = \dfrac{\mu_0 \times I}{2\pi \times r}\)
\(\text{Find}\ B\ \text{when}\ I → \dfrac{I}{2},\ \text{and}\ r → 3r:\)
\(B_{\text{new}}\) | \(=\dfrac{\mu_0 \times \frac{I}{2}}{2\pi \times 3r}\) | |
\(=\dfrac{1}{6} \times \dfrac{\mu_0 \times I}{2\pi \times r}\) | ||
\(=\dfrac{1}{6} \times 2 \times 10^{-3}\) | ||
\(=3.33 \times 10^{-4}\ \text{T}\) |
Two long, straight current-carrying wires, \(\text{P}\) and \(\text{Q}\), are parallel, as shown below. The current in the wires is the same in magnitude and opposite in direction. Figure 2b shows the wires as viewed from above. On Figure 2b, sketch the magnetic field around the wires, showing the direction of the magnetic field. Use at least five field lines. (3 marks) --- 0 WORK AREA LINES (style=blank) --- → Use the right-hand grip rule to determine the directions of the magnetic fields surrounding the currents, remembering that an \(X\) means “into the page” and the dot means “out of the page”. → As the distance from the wire increases the field strength decreases according to the inverse square law and therefore does not decrease at a linear rate (as seen in the diagram by the greater distances between field lines the further from the wires).
Some physics students are conducting an experiment investigating both electrostatic and gravitational forces. They suspend two equally charged balls, each of mass 4.0 g, from light, non-conducting strings suspended from a low ceiling. The charged balls repel each other with the strings at an angle of 60°, as shown in Figure 1. There are three forces acting on each ball: --- 0 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. → The sum of the forces must add to zero as seen in the triangle below. → \(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\) a. → The sum of the forces must add to zero as seen in the triangle below. → \(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\) c. Using the same triangle as part (b):
b. System is in an equilibrium state:
\(\sin\,60^{\circ}\)
\(=\dfrac{3.92 \times 10^{-2}}{T}\)
\(T\)
\(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}\)
\(=4.5 \times 10^{-2}\ \text{N}\)
c. Using the same triangle as part (b):
\(\tan\,60^{\circ}\)
\(=\dfrac{3.92 \times 10^{-2}}{F_E}\)
\(F_E\)
\(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}\)
\(=2.3 \times 10^{-2}\ \text{N}\)
b. System is in an equilibrium state:
\(\sin\,60^{\circ}\)
\(=\dfrac{3.92 \times 10^{-2}}{T}\)
\(T\)
\(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}\)
\(=4.5 \times 10^{-2}\ \text{N}\)
\(\tan\,60^{\circ}\)
\(=\dfrac{3.92 \times 10^{-2}}{F_E}\)
\(F_E\)
\(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}\)
\(=2.3 \times 10^{-2}\ \text{N}\)
A ray of monochromatic light is incident on a triangular glass prism with a refractive index of 1.52 . The ray is perpendicular to the side \(\text{AB}\) of the glass prism, as shown in Figure 14. The ray of light travels through the glass prism before reaching side \(\text{AC}\). --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. 41° b. As the angle of incidence 45° is greater than the critical angle 41°, the total internal reflection will occur on side \(\text{AC}\). b. Total internal reflection: → The angle of incidence 45° is greater than the critical angle 41°. → Therefore, total internal reflection will occur on side \(\text{AC}\).
a.
\(\sin \theta_c\)
\(=\dfrac{n_2}{n_1}\)
\(\theta_c\)
\(=\sin^{-1}(\dfrac{n_2}{n_1})\)
\(=\sin^{-1}(\dfrac{1.0}{1.52})\)
\(=41^{\circ}\)
A guitar string of length 0.75 m and fixed at both ends is plucked and a standing wave is produced. The envelope of the standing wave is shown in Figure 13.
The speed of the wave along the string is 393 m s\( ^{-1}\).
--- 2 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
a. 262 Hz
b. Standing wave:
→ When waves encounter fixed ends, they reflect.
→ If the string’s length is a multiple of half the wavelength, the reflected wave combines with the original wave, resulting in interference that forms a standing wave pattern
a. \(f=\dfrac{v}{\lambda}=\dfrac{393}{1.5}=262\ \text{Hz}\)
b. Standing wave:
→ When waves encounter fixed ends, they reflect.
→ If the string’s length is a multiple of half the wavelength, the reflected wave combines with the original wave, resulting in interference that forms a standing wave pattern
Maia is at a skatepark. She stands on her skateboard as it rolls in a straight line down a gentle slope at a constant speed of 3.0 m s\(^{-1}\), as shown in Figure 8. The slope is 5° to the horizontal. The combined mass of Maia and the skateboard is 65 kg. --- 0 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- Near the bottom of the ramp, Maia takes hold of a large pole and comes to a complete rest while still standing on the skateboard. Maia and the skateboard now have no momentum or kinetic energy. --- 3 WORK AREA LINES (style=lined) --- a. b. 55.5 N c. Momentum: → Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible. Kinetic energy: → The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down. a. b. Total frictional forces: → Constant speed means that the force down the slope of the incline is equal to the sum of the frictional forces acting on Maia and the skateboard. c. Momentum: → Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible. Kinetic energy: → The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.
\(F_{\text{down slope}}\)
\(=F_f\)
\(=mg\, \sin\, \theta\)
\(=65 \times 9.8 \times \sin\,5^{\circ}\)
\(=55.5\ \text{N}\)
The diagram below shows the force versus time graph of the force on a tennis ball when it is hit by a tennis racquet. The tennis ball is stationary when the tennis racquet first comes into contact with the ball.
Which one of the following is closest to the impulse experienced by the tennis ball as it is hit by the tennis racquet?
\(B\)
→ The impulse is equal to the area under the curve \((I=Ft)\).
→ By splitting the areas into whole number of squares, there are about 10 squares.
→ Each square is \(50\ \text{N} \times 0.01\ \text{s}=0.5\ \text{N s}\)
→ Hence the impulse experienced by the tennis ball \(\approx 5.0\ \text{N s}\).
\(\Rightarrow B\)
The diagram below shows two charges, \(Q_1\) and \(Q_2\), separated by a distance, \(d\).
There is a force, \(F\), acting between the two charges.
Which one of the following is closest to the magnitude of the force acting between the two charges if both \(d\) and the charge on \(Q_1\) are halved?
\(C\)
\(\text{Initially,}\ \ F=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{Q_1Q_2}{d^2}\)
\(\text{After}\ \ Q_1\ \ \Rightarrow \ \ \dfrac{Q_1}{2}, \ \text{and}\ \ d\ \ \Rightarrow \ \ \dfrac{d}{2}:\)
\(F_{\text{new}}\) | \(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{\frac{Q_1}{2}Q_2}{(\frac{d}{2})^2}\) | |
\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{\frac{Q_1}{2}Q_2}{\frac{d^2}{4}}\) | ||
\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{4Q_1Q_2}{2d^2}\) | ||
\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{2Q_1Q_2}{d^2}\) | ||
\(=2F\) |
\(\Rightarrow C\)
Standing waves are formed on a string of length 4.0 m that is fixed at both ends. The speed of the waves is 240 m s\(^{-1}\). --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. 30 Hz b. 60 Hz c. Standing waves: → Produced when a wave travels down a string and is reflected back on itself such that the superposition of the two waves produce an interference pattern to form a standing wave. → The two waves must be travelling in opposite directions with the same frequency, wavelength and amplitude. a. Lowest frequency resonance: → Occurs at the maximum wavelength. The maximum wavelength is 8 metres since the half wavelength is the length of the string (4m). \(f=\dfrac{v}{\lambda}=\dfrac{240}{8}=30\ \text{Hz}\) b. When \(\lambda = 4: \) \(f=\dfrac{v}{\lambda}=\dfrac{240}{4}=60\ \text{Hz}\) c. Standing waves: → Produced when a wave travels down a string and is reflected back on itself such that the superposition of the two waves produce an interference pattern to form a standing wave. → The two waves must be travelling in opposite directions with the same frequency, wavelength and amplitude.
The enthalpies of formation for a number of chemical reactions are as follows:
\(\ce{C6H12O6(s)}\) \(\Delta H^{\circ}_f = -1271\ \text{kJmol}^{-1}\)
\(\ce{C2H5OH(aq)}\) \(\Delta H^{\circ}_f = -277.7\ \text{kJmol}^{-1}\)
\(\ce{CO2(g)}\) \(\Delta H^{\circ}_f = -393.5\ \text{kJmol}^{-1}\)
Calculate the enthalpy change for the fermentation of glucose (reaction below) using the enthalpies of formation above.
\(\ce{C6H12O6(s) \rightarrow 2C2H5OH(aq) + 2CO2(g)}\) (3 marks)
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\(\Delta H^{\circ}_{\text{reaction}} = -71.4\ \text{kJmol}^{-1}\)
\(\Sigma\ \text{Enthalpy (reactants)}\ = 1 \times -1271 = -1271\ \text{kJ}\)
\(\Sigma\ \text{Enthalpy (products)}\ = 2 \times -277.7 + 2 \times -393.5 = -1342.4\ \text{kJ}\)
\(\Delta H^{\circ}_{\text{reaction}}\) | \(= \Delta H^{\circ} (\text{products})-\Delta H^{\circ} (\text{reactants}) \) | |
\(=-1342.4-(-1271)\) | ||
\(= -71.4\ \text{kJ mol}^{-1}\) |
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a. → The Neolithic revolution refers to the beginning of agriculture
→ It occurred about 10’000 years ago and includes the first instances of cultivating crops and domesticating animals.
b. The Neolithic Revolution
→ The Neolithic revolution heavily influenced the evolution of the human race, as well as the impacted ecosystems and related flora and fauna.
→ One immediate impact was an increase in human populations. The beginning of agriculture helped remove a gathering lifestyle and provided an abundance of available food, leading to the first instances of villages and towns.
→ Another impact is soil erosion; the removal of the top soil layer due to removing deep-rooted vegetation. The top layer of soil provides many organic matter with essential minerals and its removal can majorly disrupt entire ecosystems. This also developed into a major problem in todays agriculture, deforestation.
a. → The Neolithic revolution refers to the beginning of agriculture
→ It occurred about 10’000 years ago and includes the first instances of cultivating crops and domesticating animals.
b. The Neolithic Revolution
→ The Neolithic revolution heavily influenced the evolution of the human race, as well as the impacted ecosystems and related flora and fauna.
→ One immediate impact was an increase in human populations. The beginning of agriculture helped remove a gathering lifestyle and provided an abundance of available food, leading to the first instances of villages and towns.
→ Another impact is soil erosion; the removal of the top soil layer due to removing deep-rooted vegetation. The top layer of soil provides many organic matter with essential minerals and its removal can majorly disrupt entire ecosystems. This also developed into a major problem in todays agriculture, deforestation.
Three unknown metals are reacted with dilute \(\ce{HCl(aq)}\) and the following observations are made:
\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \textit{Metal} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \textit{Observations} \\
\hline
\rule{0pt}{2.5ex} \text{A} \rule[-1ex]{0pt}{0pt} & \text{No observable reaction} \\
\hline
\rule{0pt}{2.5ex} \text{B} \rule[-1ex]{0pt}{0pt} & \text{Slow bubbling} \\
\hline
\rule{0pt}{2.5ex} \text{C} \rule[-1ex]{0pt}{0pt} & \text{Fast, abrupt bubbling} \\
\hline
\end{array}
You are told that the metals in question are Magnesium, Platinum and Zinc.
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i. → The metals \(\text{A}\), \(\text{B}\) and \(\text{C}\) differ by their relative reactivity.
→ Magnesium is the most reactive as it is an alkali earth metal (Metal \(\text{C}\)).
→ Zinc and Platinum are both less reactive than magnesium. Zinc however exists higher on the reactivity series of metals than Platinum which is the least reactive of the three metals.
→ Therefore, metal \(\text{B}\) is zinc and metal \(\text{A}\) is platinum.
ii. → A single displacement reaction.
iii. Include one of the following:
\(\ce{Mg(s) + 2HCl(aq) \rightarrow MgCl2(aq) + H2(g) }\)
\(\ce{Zn(s) + 2HCl(aq) \rightarrow ZnCl2(aq) + H2(g) }\)
i. → The metals \(\text{A}\), \(\text{B}\) and \(\text{C}\) differ by their relative reactivity.
→ Magnesium is the most reactive as it is an alkali earth metal (Metal \(\text{C}\)).
→ Zinc and Platinum are both less reactive than magnesium. Zinc however exists higher on the reactivity series of metals than Platinum which is the least reactive of the three metals.
→ Therefore, metal \(\text{B}\) is zinc and metal \(\text{A}\) is platinum.
ii. → A single displacement reaction.
iii. Include one of the following:
\(\ce{Mg(s) + 2HCl(aq) \rightarrow MgCl2(aq) + H2(g) }\)
\(\ce{Zn(s) + 2HCl(aq) \rightarrow ZnCl2(aq) + H2(g) }\)
Climate change is altering the Earth’s habitats and ecosystems, putting many species at risk of global population reductions or extinction.
Assess this statement. (5 marks)
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→ Climate change is altering organisms environments across the globe, causing drastic population reductions and even extinctions of entire species.
→ One reason for sudden changes in populations is the increase in devastating and extreme weather events because of climate change. Storms, fires and large timeframes of drought initiate large death tolls in communities, eliminating individuals and drastically changing the gene pool.
→ Climate change also has many gradual changes on the environment, including a steadily increasing temperature and changes in precipitation patterns. Many organisms cannot adapt to these changes causing many to die out and even lead to climate driven extinctions.
→ One example of a changing temperature on the environment is ocean acidification. Rising temperatures have a drastic effect on marine life such as the death of many coral species and the acidification of the water makes it difficult for many crustaceans to form hard shells.
→ The ecosystems themselves also play a large role in climate change, as plants act as a climate regulator due to their ability to store carbon. However as climate change grows at an exponential rate plants cannot keep up and many also suffer to the effects of changing precipitation patterns and temperature increases and die out, resulting in less carbon storage and an increasing rate of the above effects and their consequences.
→ It is therefore accurate to state that climate change alters ecosystems worldwide causing large population declines of species and eventually climate extinctions for some.
→ Climate change is altering organisms environments across the globe, causing drastic population reductions and even extinctions of entire species.
→ One reason for sudden changes in populations is the increase in devastating and extreme weather events because of climate change. Storms, fires and large timeframes of drought initiate large death tolls in communities, eliminating individuals and drastically changing the gene pool.
→ Climate change also has many gradual changes on the environment, including a steadily increasing temperature and changes in precipitation patterns. Many organisms cannot adapt to these changes causing many to die out and even lead to climate driven extinctions.
→ One example of a changing temperature on the environment is ocean acidification. Rising temperatures have a drastic effect on marine life such as the death of many coral species and the acidification of the water makes it difficult for many crustaceans to form hard shells.
→ The ecosystems themselves also play a large role in climate change, as plants act as a climate regulator due to their ability to store carbon. However as climate change grows at an exponential rate plants cannot keep up and many also suffer to the effects of changing precipitation patterns and temperature increases and die out, resulting in less carbon storage and an increasing rate of the above effects and their consequences.
→ It is therefore accurate to state that climate change alters ecosystems worldwide causing large population declines of species and eventually climate extinctions for some.
The landing gear on an aircraft uses a hydraulic braking system. A force of 60 N is applied at the master cylinder with a piston diameter of 12 mm.
What is the force at the brake calliper with a piston diameter of 42 mm ?
\( D \)
\(\Rightarrow D \)
A portion of a roller coaster wheel sub-assembly is shown. An exploded pictorial of the wheel sub-assembly is shown. Complete an assembled sectioned front view of the wheel sub-assembly at scale 1: 2. Apply AS 1100 drawing standards. Do NOT add dimensions. (6 marks) --- 0 WORK AREA LINES (style=lined) ---