A car of mass 1200 kg travels at a constant speed up a slope which has a gradient of 1 in 10. The resistance to motion is 300 N. Calculate the maximum speed of the car if the power of the engine is 54

Mass of car, $m = 1200 \text{ kg}$
Power, $P = 54 \text{ kW} = 54000 \text{ W}$
Angle of inclination, $\theta = \sin^{-1} \left(\frac{1}{10}\right)$, so $\sin \theta = \frac{1}{10}$
Non-gravitational resistance, $R = 300 \text{ N}$
Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$
At maximum speed, the net force is zero, and power is given by $P = Fv$, where $F$ is the total force opposing motion.

The total opposing force is the sum of the gravitational component and the non-gravitational resistance.

Using the power formula $P = F_{total} v$:

The options provided are A. $20 \text{ m s}^{-1}$, B. $25 \text{ m s}^{-1}$, C. $100 \text{ m s}^{-1}$, D. $900 \text{ m s}^{-1}$.
There seems to be a discrepancy between the calculated answer and the given options. Let me recheck the calculation.
$P = 54 \text{ kW} = 54000 \text{ W}$.
$m = 1200 \text{ kg}$.
$\sin \theta = 1/10$.
$R = 300 \text{ N}$.
$g = 10 \text{ m s}^{-2}$.
Force due to gravity down the incline: $F_g = mg \sin \theta = 1200 \times 10 \times (1/10) = 1200 \text{ N}$.
Total resistive force: $F_{total} = F_g + R = 1200 \text{ N} + 300 \text{ N} = 1500 \text{ N}$.
Maximum speed: $v = P / F_{total} = 54000 \text{ W} / 1500 \text{ N} = 36 \text{ m s}^{-1}$.

Given the options, none of them match $36 \text{ m s}^{-1}$. It's possible there's a typo in the question's values or the options. However, based on the provided information, the calculated maximum speed is $36 \text{ m s}^{-1}$. If I had to choose the closest option, it would be B. $25 \text{ m s}^{-1}$ or A. $20 \text{ m s}^{-1}$, but $36 \text{ m s}^{-1}$ is the direct calculation. I will state the calculated answer.
The maximum speed of the car is $\boxed{36 \text{ m s}^{-1}}$.

Q48.

Probability of landing head-up, $P(H) = \frac{1}{4}$.
Number of tosses, $n = 3$.

The probability of exactly $k$ successes in $n$ trials is given by $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Here, "success" is getting a tail, so $p = P(T) = \frac{3}{4}$. We want $k=1$ tail in $n=3$ tosses.

The correct option is A.
The probability of getting exactly one tail is $\boxed{\text{A. } \frac{9}{64}}$.

Q49.

Particle 1: $m_1 = 2 \text{ kg}$, $(x_1, y_1) = (3, 1)$
Particle 2: $m_2 = 3 \text{ kg}$, $(x_2, y_2) = (8, 6)$

The position of their center of mass is $(6, 4)$.
The correct option is A.
The position of their center of mass is $\boxed{\text{A. } (6,4)}$.

Q50.

Force, $F = 14 \text{ N}$
Mass, $m = 70 \text{ kg}$
Initial speed, $u = 3 \text{ m s}^{-1}$
Final speed, $v = 7 \text{ m s}^{-1}$

The work done by the force is equal to the change in kinetic energy.

The correct option is D.
The distance travelled is $\boxed{\text{D. } 100 \text{ m}}$.