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E = 60%, L = 30N, E = 20N M.A = \(\frac{\text{L}}{\text{E}}\) = \(\frac{30}{20}\) = 1.5

Average speed = \(\frac{\text{total distance}}{\total time taken}}\) d = 1. 0km ⇒ 1000m t = 5minutes ⇒ 5 x 60 = 300 seconds Average speed = \(\frac{1000}{300}\) = 3.3m/s

Uniform speed = 120km/hr, d =?, t = 4minutes ⇒ \(\frac{4}{60}\) = 0.0667hrs Distance = speed x time D = 120 x 0.0667 = 8km.

Answer: B For equilibrium, clockwise moment = anticlockwise moment about the pivot. clockwise distance from pivot: 50cm - 10cm = 40cm anticlockwise distance from pivot: 85cm - 10cm = 75cm Applying the principle of moments W x distance(w) = T x distance(T) 100 x 40 = T x 75 T = \(\frac{ 4000}{75}\) ≈ 53.33N

T x 80 + 15 x 10 = W x 50 80T + 150 = 50W - - -- - - - - -(1) T + 15 = W - - - - - - - - - - - (2) 80T + 150 = 50(T + 15) 80T + 150 = 50T + 750 30T = 750 - 150 30T = 600 T = 20N The closest option is 19.2N

In a vacuum (where air resistance is ignored), the maximum horizontal range is achieved at an angle of 45º. From R = \(\frac{u^2sin2\theta}{\text{g}}\) R\(_{max}\) = \(\frac{u^2sin2(45)}{\text{g}}\) = \(\frac{u^2}{\text{g}}\) \(\therefore\) \(\theta\) = 45º

C. 6 kg Initial velocity \( u = 4 \, \text{m/s} \), Acceleration \( a = 5 \, \text{m/s}^2 \), Time \( t = 2 \, \text{s} \) Final velocity: \(v = u + at = 4 + 5 \times 2 = 14 \, \text{m/s}\) Change in momentum (impulse): \(\Delta p = m(v - u) = 60\) Substituting for \( v \): \(m(14 - 4) = 60\) Simplifying: \(m \times 10 = 60\) Thus, \(m = 6 \, \text{kg}\)

A. Q In a simple pendulum, kinetic energy is maximum at the lowest point (equilibrium position) where potential energy is minimum and speed is maximum.Here, Q is the control point (mean/equilibrium position), so maximum kinetic energy occurs at Q. At extremes P and S, kinetic energy is zero (velocity = 0). At R, it is between the extreme and the equilibrium.

The frequency \( f \) of a simple pendulum is: \(f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \quad \text{where } L = 1.2 \, \text{m}, \, g = 10 \, \text{m/s}^2, \, \pi \approx \frac{22}{7}\) Calculate: \(\frac{g}{L} = \frac{10}{1.2} = \frac{25}{3}, \quad \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}\) \(f = \frac{1}{2 \times \frac{22}{7}} \times \frac{5}{\sqrt{3}} = \frac{35}{44\sqrt{3}}\) Approximating \( \sqrt{3} \approx 1.732 \): \(44 \times 1.732 \approx 76.208 \quad \Rightarrow \quad \frac{35}{76.208} \approx 0.459 \, \text{Hz}\) Thus, the answer is \( D. \, 0.5 \) Hz.

Period of oscillation = \(\frac{\text{ time}}{\text{no. of oscillations}}\) = \(\frac{38}{20}\) = 1.9s

Linear speed = \(\omega\) r \(\omega\) = \(\frac{\text{v}}{\text{r}}\) \(\omega\) = \(\frac{10}{4}\) = 2.5 rads\(^{-1}\)

The frequency (f) of a simple pendulum is the number of oscillations it completes per second and is given by the formula: f = \(\frac{1}{2\pi}\)\(\sqrt{\frac{\text{g}}{\text{l}}}\) where l is the length of the string and g is the acceleration due to gravity. Because the length l is in the denominator, the frequency is inversely proportional to the square root of the length. Therefore, increasing the length results in a slower swing rate, effectively reducing the frequency.

Answer: B. 0.015 (Net F = ma = 40\times0.3 = 12 N; friction = 18 - 12 = 6 N; \mu = \(\frac{\text{friction}}{\text{mg}}\) = \(\frac{6}{(40\times10)}\) = 0.015.)

Frictional force is the tangential force opposing an object's free sliding on an adjacent surface. It originates from surface irregularities and intermolecular attractions between materials, resisting relative motion or its initiation.

Given: g = 9.8 m/s\(^2\), a = 10 m/s, v = 0( come to rest), s = 25m v\(^2\) = u\(^2\) + 2as - 100 = 2 x 25 x a a = \(\frac{-100}{50}\) = - 2 deceleration is caused by kinetic friction But, \(\mu\) = \(\frac{\text{f}}{\text{R}}\) = \(\frac{\text{ma}}{\text{mg}}\) = \(\frac{-2}{9.8}\) ≈ 0.2.