Question

1. (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/s², 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 reference 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 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

Explanation:

Problem 8(a)(ii)

Step1: Apply Newton's second law

According to Newton's second - law $F_{net}=ma$. The applied force $F = 35N$, and the mass $m = 25kg$, and the acceleration $a=1.5m/s^{2}$. The net force $F_{net}=F - f$, where $f$ is the frictional force. So $F - f=ma$.

Step2: Rearrange the formula to find frictional force

We can rewrite the equation $F - f=ma$ as $f=F - ma$. Substitute $F = 35N$, $m = 25kg$ and $a = 1.5m/s^{2}$ into the formula. $f=35-25\times1.5$.

Step3: Calculate the value of frictional force

First, calculate $25\times1.5 = 37.5N$. Then $f=35 - 37.5=- 2.5N$. The negative sign just indicates that the frictional force acts in the opposite direction of the motion. The magnitude of the frictional force is $2.5N$.

Step1: Calculate the load

The load $L$ is the weight of the car. Using $L = mg$, where $m = 8000kg$ and $g = 9.8m/s^{2}$. So $L=8000\times9.8 = 78400N$.

Step2: Calculate the mechanical advantage (MA) and velocity ratio (VR) for a screw - jack

The velocity ratio of a screw - jack is given by $VR=\frac{2\pi l}{p}$, where $l$ is the length of the tommy - bar and $p$ is the pitch. Given $p = 2mm=0.002m$ and $l = 25cm = 0.25m$. Then $VR=\frac{2\pi\times0.25}{0.002}=\frac{0.5\pi}{0.002}=250\pi$.

Step3: Use the efficiency formula $\eta=\frac{MA}{VR}$

We know that $\eta = 85\%=0.85$, and $MA=\frac{L}{E}$, where $E$ is the effort. From $\eta=\frac{MA}{VR}$, we can rewrite it as $MA=\eta\times VR$. And since $MA=\frac{L}{E}$, then $E=\frac{L}{\eta\times VR}$.

Step4: Substitute the values to find the effort

Substitute $L = 78400N$, $\eta = 0.85$ and $VR = 250\pi$ into the formula. $E=\frac{78400}{0.85\times250\pi}\approx117.7N$.

Step1: Use the principle of heat exchange $Q_{lost}=Q_{gained}$

The heat lost by the iron rod is $Q_{iron}=m_{iron}c_{iron}(T_{1}-T_{2})$, and the heat gained by the water is $Q_{water}=m_{water}c_{water}(T_{2}-T_{3})$. Let $m_{iron}=2.5kg$, $T_{1}=250^{\circ}C$, $T_{2}=72^{\circ}C$, $T_{3}=33^{\circ}C$. The specific heat capacity of iron $c_{iron}=460J/(kg\cdot^{\circ}C)$ and the specific heat capacity of water $c_{water}=4200J/(kg\cdot^{\circ}C)$.

Step2: Set up the equation and solve for $m_{water}$

From $Q_{lost}=Q_{gained}$, we have $m_{iron}c_{iron}(T_{1}-T_{2})=m_{water}c_{water}(T_{2}-T_{3})$. Then $m_{water}=\frac{m_{iron}c_{iron}(T_{1}-T_{2})}{c_{water}(T_{2}-T_{3})}$.

Step3: Substitute the values and calculate

Substitute $m_{iron}=2.5kg$, $c_{iron}=460J/(kg\cdot^{\circ}C)$, $T_{1}=250^{\circ}C$, $T_{2}=72^{\circ}C$, $c_{water}=4200J/(kg\cdot^{\circ}C)$ and $T_{3}=33^{\circ}C$ into the formula.
\[

$$\begin{align*} m_{water}&=\frac{2.5\times460\times(250 - 72)}{4200\times(72 - 33)}\\ &=\frac{2.5\times460\times178}{4200\times39}\\ &=\frac{2.5\times460\times178}{163800}\\ &\approx1.01kg \end{align*}$$

\]

Answer:

$2.5N$

Problem 8(c)(i)