8. (a) (i) draw a pulley system that has a velocity ratio of 4.
(ii) a body of mass 25 kg is pulled over a rough surface with a 35 n force. if the object accelerated at a rate of 1.5 m s⁻², calculate the frictional force acting on the object and the surface.
5 marks
(b) define the following terms as applied to machines:
(i) velocity ratio:
(ii) efficiency.
4 marks
(c) (i) a screw jack of pitch 2 mm is to be used to lift a car of mass 8000 kg. the length of the tommy bar of the jack is 25 cm. calculate the effort that would have been required to attain an efficiency of 85 %.
(ii) explain why the efficiency of a machine is always less than 100 %.
6 marks
9. (a) (i) define the term thermal conductivity.
(ii) give one difference between latent heat of fusion and latent heat of vaporization.
4 marks
(b) (i) an iron rod of mass 2.5 kg at 250 °c is dropped into some quantity of water initially at 33 °c. what would be the mass of the water when the temperature is at 72 °c?
(ii) a piece of ice at −15 °c is subjected to heat until the ice changes to steam at 100 °c. sketch a heating curve to illustrate the changes in temperature during the process.
7 marks
(c) (i) state two effects of heat on a substance.
(ii) by how much should water of temperature −25 °c be increased to obtain its freezing point temperature?
4 marks
10. (a) (i) state two characteristics of resonance.
(ii) mention the effects of temperature and pressure on the speed of sound.
(iii) list the types of resonance as applied to wave.
7 marks
(b) a referee leaning on a wall blows his whistle towards another wall 50 m away. he hears the second echo 2 seconds later. calculate the velocity of the sound of the whistle in air.
speed of sound in air = 330 m s⁻¹
3 marks
(c) (i) draw a diagram illustrating convergent and divergent beams of light.
(ii) the critical angle for glass in air is 39°. calculate the refractive index of the glass.
5 marks

Question

Accepted Answer

### 8 (a)(ii) Solution:
# Explanation:
## Step 1: Recall Newton's Second Law
Newton's second law is $ F_{net} = ma $, where $ F_{net} $ is the net force, $ m $ is mass, and $ a $ is acceleration. The net force is also $ F_{applied}-F_{friction} $, so $ F_{applied}-F_{friction}=ma $.
## Step 2: Calculate the net force
Given $ m = 25\space kg $ and $ a = 1.5\space m/s^2 $, the net force $ F_{net}=ma = 25\times1.5 = 37.5\space N $.
## Step 3: Solve for frictional force
We know $ F_{applied}=35\space N $ and $ F_{applied}-F_{friction}=F_{net} $. Rearranging gives $ F_{friction}=F_{applied}-F_{net} $. Wait, but $ F_{applied}(35\space N) $ is less than $ F_{net}(37.5\space N) $? That can't be. Wait, no—wait, maybe I mixed up. Wait, the net force should be $ F_{applied}-F_{friction}=ma $, but if the object is accelerating, the applied force must be greater than friction. Wait, maybe the mass is 25 kg, so weight is $ mg $, but no, the force is horizontal. Wait, no, the formula is correct: $ F_{net}=F_{applied}-F_{friction}=ma $. So $ F_{friction}=F_{applied}-ma $. Wait, but $ 35 - (25\times1.5)=35 - 37.5=- 2.5 $. That doesn't make sense. Wait, maybe I had the formula wrong. The correct formula is $ F_{net}=ma $, and $ F_{net}=F_{applied}-F_{friction} $ (if applied force is in the direction of motion, friction opposite). So if the object is accelerating, $ F_{applied}>F_{friction} $, so $ F_{friction}=F_{applied}-ma $. But here, $ 35 - 37.5 $ is negative, which would mean friction is in the direction of motion, which is impossible. Wait, maybe the acceleration is deceleration? Or maybe the mass is 2.5 kg? Wait, the problem says 25 kg. Wait, maybe I made a mistake. Wait, let's check again. $ m = 25\space kg $, $ a = 1.5\space m/s^2 $, so $ ma = 25\times1.5 = 37.5\space N $. The applied force is 35 N, which is less than 37.5 N. That would mean the net force is negative, so the object is decelerating, but the problem says "accelerated at a rate of $ 1.5\space m/s^2 $". So maybe there's a typo, but assuming the problem is correct, maybe the formula is $ F_{friction}-F_{applied}=ma $ (if friction is greater, but then acceleration would be negative). Wait, no—Newton's second law is $ \sum F = ma $. So if the object is moving, and there's an applied force $ F $ and friction $ f $, then $ F - f = ma $ (if $ F>f $, acceleration in direction of $ F $) or $ f - F = ma $ (if $ f>F $, acceleration opposite to $ F $, i.e., deceleration). But the problem says "accelerated at a rate of $ 1.5\space m/s^2 $", so $ a $ is positive in the direction of $ F $, so $ F - f = ma $. But $ F = 35\space N $, $ ma = 37.5\space N $, so $ 35 - f = 37.5 \implies f = 35 - 37.5=- 2.5\space N $. Negative friction doesn't make sense. So maybe the mass is 2.5 kg? Let's try that. If $ m = 2.5\space kg $, then $ ma = 2.5\times1.5 = 3.75\space N $, so $ f = 35 - 3.75 = 31.25\space N $. That makes sense. Maybe a typo in the problem, mass is 2.5 kg. Assuming that, let's proceed.

Wait, the original problem says 25 kg. Maybe I misread. Let's check again: "A body of mass 25 kg is pulled over a rough surface with a 35 N force. If the object accelerated at a rate of $ 1.5\space m/s^2 $, calculate the frictional force acting on the object and the surface."

So according to Newton's second law, $ F_{net}=ma = 25\times1.5 = 37.5\space N $. The net force is also $ F_{applied}-F_{friction} $, so $ 35 - F_{friction}=37.5 \implies F_{friction}=35 - 37.5=- 2.5\space N $. Negative friction implies that the friction force is in the direction of the applied force, which would mean the surface is helping the motion, which is not possible for friction (friction opposes relative motion). So there must be an error in the problem, but assuming that maybe the acceleration is $ 0.5\space m/s^2 $, or the force is 40 N. Alternatively, maybe the mass is 2.5 kg. Let's assume the mass is 2.5 kg (maybe a typo, 2.5 instead of 25). Then $ ma = 2.5\times1.5 = 3.75\space N $, so $ F_{friction}=35 - 3.75 = 31.25\space N $.

But since the problem states 25 kg, maybe the intended solution is to use $ F_{friction}=F_{applied}-ma $, ignoring the sign, so $ |35 - 37.5| = 2.5\space N $, but that would mean friction is in the direction of motion, which is incorrect. Alternatively, maybe the problem has a mistake, but we'll proceed with the given numbers.

# Answer:
Assuming the problem has a typo (mass should be 2.5 kg) or we consider the magnitude, the frictional force is $ \boldsymbol{2.5\space N} $ (but this is physically inconsistent with the given acceleration and force; likely a typo in mass, correct mass should be 2.5 kg, leading to $ 31.25\space N $).

### 8 (c)(i) Solution:
# Explanation:
## Step 1: Recall the formula for the mechanical advantage (MA) of a screw jack
The mechanical advantage of a screw jack is given by $ MA=\frac{2\pi L}{p} $, where $ L $ is the length of the tommy bar, and $ p $ is the pitch.
## Step 2: Recall the formula for efficiency ($ \eta $)
Efficiency $ \eta=\frac{MA}{VR}\times100\% $, where $ VR $ is the velocity ratio (for a screw jack, $ VR = \frac{2\pi L}{p} $, so $ MA=\eta\times VR/100 $). Also, $ MA=\frac{Load}{Effort} $, where $ Load = mg $.
## Step 3: Calculate the load
Given $ m = 8000\space kg $, $ g = 9.8\space m/s^2 $, so $ Load = mg = 8000\times9.8 = 78400\space N $.
## Step 4: Calculate the velocity ratio (VR)
Pitch $ p = 2\space mm = 0.002\space m $, length $ L = 25\space cm = 0.25\space m $. $ VR=\frac{2\pi L}{p}=\frac{2\times\pi\times0.25}{0.002}=\frac{0.5\pi}{0.002}=250\pi\approx785.4 $.
## Step 5: Calculate the mechanical advantage (MA)
Efficiency $ \eta = 85\% = 0.85 $. $ MA=\eta\times VR = 0.85\times785.4\approx667.6 $.
## Step 6: Calculate the effort (E)
Since $ MA=\frac{Load}{Effort} $, $ Effort=\frac{Load}{MA}=\frac{78400}{667.6}\approx117.4\space N $.

# Answer:
The effort required is approximately $ \boldsymbol{117\space N} $ (or more precisely $ \approx117.4\space N $).

### 8 (c)(ii) Explanation:
# Brief Explanations:
The efficiency of a machine is always less than 100% because of energy losses. These losses occur due to friction between moving parts (converting energy to heat), air resistance, and the work done in overcoming the weight of moving parts (like in pulleys, the weight of the pulleys themselves). Some energy is also lost as sound or due to deformation of components. Thus, the output work (useful work) is always less than the input work, making efficiency less than 100%.
# Answer:
Efficiency < 100% due to energy losses (friction, air resistance, etc.).

### 9 (a)(i) Explanation:
# Brief Explanations:
Thermal conductivity is the ability of a material to conduct heat. It is defined as the rate of heat transfer ($ Q $) per unit time through a unit cross - sectional area ($ A $) of the material, per unit temperature gradient ($ \frac{\Delta T}{\Delta x} $) across the material. Mathematically, $ k=\frac{Q\Delta x}{A\Delta T t} $, where $ k $ is the thermal conductivity, $ t $ is time, $ \Delta x $ is the thickness of the material, and $ \Delta T $ is the temperature difference across the material.
# Answer:
Thermal conductivity is the rate of heat transfer through a unit area, per unit temperature gradient, per unit time.

### 9 (a)(ii) Explanation:
# Brief Explanations:
Latent heat of fusion is the heat energy required to change a unit mass of a substance from solid to liquid state at its melting point (without a change in temperature). Latent heat of vaporization is the heat energy required to change a unit mass of a substance from liquid to gaseous state at its boiling point (without a change in temperature). One difference is that latent heat of vaporization is generally greater than latent heat of fusion for the same substance (e.g., for water, latent heat of fusion is about $ 334\space kJ/kg $ and latent heat of vaporization is about $ 2260\space kJ/kg $). Another difference is that fusion involves a solid - liquid phase change, while vaporization involves a liquid - gas phase change.
# Answer:
Latent heat of fusion is for solid - liquid change, vaporization for liquid - gas; vaporization latent heat is larger (for same substance).

### 9 (b)(i) Solution:
# Explanation:
## Step 1: Recall the principle of calorimetry
The heat lost by the iron rod is equal to the heat gained by the water, i.e., $ Q_{lost}=Q_{gained} $.
## Step 2: Write the heat equations
Heat lost by iron: $ Q_{iron}=m_{iron}c_{iron}(T_{iron,initial}-T_{final}) $, where $ m_{iron}=2.5\space kg $, $ c_{iron}=460\space J/(kg\cdot^{\circ}C) $, $ T_{iron,initial}=250^{\circ}C $, $ T_{final}=72^{\circ}C $.

Heat gained by water: $ Q_{water}=m_{water}c_{water}(T_{final}-T_{water,initial}) $, where $ c_{water}=4200\space J/(kg\cdot^{\circ}C) $, $ T_{water,initial}=33^{\circ}C $, $ T_{final}=72^{\circ}C $.
## Step 3: Set $ Q_{iron}=Q_{water} $
$ 2.5\times460\times(250 - 72)=m_{water}\times4200\times(72 - 33) $
## Step 4: Calculate the left - hand side
$ 2.5\times460\times178 = 2.5\times81880 = 204700\space J $
## Step 5: Calculate the right - hand side
$ m_{water}\times4200\times39=m_{water}\times163800 $
## Step 6: Solve for $ m_{water} $
$ m_{water}=\frac{204700}{163800}\approx1.25\space kg $

# Answer:
The mass of the water is approximately $ \boldsymbol{1.25\space kg} $.

### 9 (b)(ii) Explanation:
# Brief Explanations:
The heating curve for ice at $ - 15^{\circ}C $ to steam at $ 100^{\circ}C $ has the following stages:
1. **Stage 1: Heating ice from $ - 15^{\circ}C $ to $ 0^{\circ}C $ (solid phase, temperature increases):** The temperature of the ice rises linearly with time (since heat is being added, and the specific heat capacity of ice is constant). The slope of this part of the graph depends on the specific heat capacity of ice ($ c_{ice} $).
2. **Stage 2: Melting of ice at $ 0^{\circ}C $ (solid - liquid phase change, temperature constant):** During this stage, heat is added but the temperature remains at $ 0^{\circ}C $ as the ice melts into water. This is a horizontal line on the temperature - time graph, and the length of this line depends on the latent heat of fusion of ice.
3. **Stage 3: Heating water from $ 0^{\circ}C $ to $ 100^{\circ}C $ (liquid phase, temperature increases):** The temperature of the water rises linearly with time, with a slope depending on the specific heat capacity of water ($ c_{water} $). Since $ c_{water}>c_{ice} $, the slope of this part is less steep than the slope of the first stage (if the rate of heat addition is constant).
4. **Stage 4: Vaporization of water at $ 100^{\circ}C $ (liquid - gas phase change, temperature constant):** Heat is added, but the temperature remains at $ 100^{\circ}C $ as water turns into steam. This is another horizontal line on the graph, and its length depends on the latent heat of vaporization of water (which is much longer than the length of the melting stage line, since latent heat of vaporization is greater than latent heat of fusion for water).

The x - axis is time, and the y - axis is temperature.
# Answer:
Heating curve has 4 stages: heat ice to $ 0^{\circ}C $ (temp rise), melt ice (temp constant), heat water to $ 100^{\circ}C $ (temp rise), vaporize water (temp constant).

### 9 (c)(i) Explanation:
# Brief Explanations:
Two effects of heat on a substance are:
1. **Change in temperature:** When heat is added to a substance (without a phase change), its temperature increases. For example, heating a metal rod increases its temperature. When heat is removed, the temperature decreases.
2. **Change in state (phase change):** Heat can cause a substance to change from one state to another. For example, adding heat to ice (solid) at $ 0^{\circ}C $ causes it to melt into water (liquid) (latent heat of fusion), and adding heat to water at $ 100^{\circ}C $ causes it to vaporize into steam (gas) (latent heat of vaporization). Another example is the deposition of gas to solid (like frost formation) when heat is removed.
# Answer:
1. Change in temperature. 2. Change in state (phase change).

### 9 (c)(ii) Solution:
# Explanation:
## Step 1: Recall the freezing point of water
The freezing point of water is $ 0^{\circ}C $.
## Step 2: Calculate the temperature increase
The initial temperature of water is $ - 25^{\circ}C $. The temperature increase $ \Delta T=T_{final}-T_{initial}=0 - (- 25)=25^{\circ}C $.

# Answer:
The temperature should be increased by $ \boldsymbol{25^{\circ}C} $.

### 10 (a)(i) Explanation:
# Brief Explanations:
Two characteristics of resonance are:
1. **Large amplitude:** When a system is in resonance, it oscillates with a much larger amplitude compared to the amplitude of the driving force when the frequency of the driving force is not equal to the natural frequency of the system.
2. **Frequency matching:** Resonance occurs when the frequency of the external (driving) force is equal to the natural frequency of the vibrating system.

# Answer

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QUESTION IMAGE

Question

Explanation:

8 (a)(ii) Solution:

Step 1: Recall Newton's Second Law

Step 2: Calculate the net force

Step 3: Solve for frictional force

Step 1: Recall the formula for the mechanical advantage (MA) of a screw jack

Step 2: Recall the formula for efficiency (\( \eta \))

Step 3: Calculate the load

Step 4: Calculate the velocity ratio (VR)

Step 5: Calculate the mechanical advantage (MA)

Step 6: Calculate the effort (E)

Answer:

8 (c)(i) Solution: