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Centripetal Force (F\(_c\)) = \(\frac{mv ^2}{r}\) = \(m\omega^2r\) (Since V = \omegar) F = 0.01 x \(100^2 \times 0.2 \) F = 20 N

From principle of conservation of linear momentum, (0.05 x 200) + (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic). 10 + 0 = V Thus V = 10m/s. Recall Kinetic Energy = \(\frac{1}{2} mv^2\) \(\therefore\) K.E = 1/2 (0.05 + 0.95) x 10\(^2\) K.E = 1/2 (1 x 100) = 50 J.

Total Energy = Kinetic Energy + Potential Energy P.E = mgh = 0.1 x 10 x 10= 10 J K.E = \(\frac{1}{2}\) MV\(^2\) = \(\frac{1}{2}\) x 0.1 x 10\(^2\) = 0.5 x 0.1 x 100 = 5J T.E = K.E + P.E = 5 + 10 = 15J

To calculate the power developed in the engine, we need to determine the work done and the time taken. 1. Calculate the final kinetic energy (KE) of the car: \(KE = \frac{1}{2}mv^2 \) where \( m = 800 \text{ kg} \) and \( v = 25 \text{ m/s} \). \( KE = \frac{1}{2} \times 800 \, \text{kg} \times (25 \, \text{m/s})^2 \) \( KE = 0.5 \times 800 \times 625 = 250000 \text{ J} \, (\text{joules})\) 2. Calculate the power (P) developed in the engine using the formula: \(P = \frac{W}{t} \) where \( W \) is the work done (in this case, equal to the kinetic energy calculated) and \( t \) is the time taken. \(P = \frac{250000 \, \text{J}}{20 \, \text{s}} = 12500 \text{ W} \, (\text{watts}) \) Thus, the power developed in the engine is 12500 watts (or 12.5 kW).

When the brakes in a car are applied, the frictional force on the tires acts in the opposite direction of the car's motion, slowing down the vehicle.

Y = \(\frac{\text{stress}}{\text{strain}}\), but stress = 10\(^7\)Nm\(^{-2}\) (given). Strain = \(\frac{\text{e}}{l_o}\). = \(\frac{(0.1005 - 0.100)}{0.1}\) = 0.005m Thus Y = \(\frac{10^7}{0.005}\) = 2.0 x 10\(^9\)Nm\(^{-2}\) (NOTE: cm → m)

R.D = \(\frac{\text{loss in weight in liquid X}}{\text{ loss in weight in water}}\) R.D = \(\frac{(10 - 7)}{(10 - 6)}\) = \(\frac{3}{4}\)

As temperature rises, the kinetic energy of the molecules increases, reducing the cohesive forces at the surface and leading to a decrease in surface tension.

At constant T, P1 V1 = P2 V2. Let P1 = P0 and V1 = V0, then solve accordingly. As pressure is varies inversely as volume according to: Boyle's law, also referred to as the Boyle–Mariotte law, or Mariotte's law, is an experimental gas law that describes how the pressure of a gas tends to decrease as the volume of the container increases. As volume is reduced by one-fifth; pressure will increase by five times.

Energy at the height = mgh This energy is converted to heat energy - mc\Deltat ∴ From the principle of energy conversion, mc\Deltat = mgh m x 450 x \Deltat = m x 10 x 20 ∴ \Deltat = \(\frac{m \times 10 \times 20}{(m \times 450)}\) = \(\frac{200}{450}\) = \(\frac{4}{9}\)^\circC

No expanded working has been attached for this question yet. The verified answer is Option A.

Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\) = \(\frac{80 - 20}{0.02}\) = \(\frac{60}{0.02}\) = 3000Km-1 = 3.0 x 103Km-1

Heat given off by the copper = MCt = 2 x 400 x (100 - 0) Heat absorbed by the ice = ML = M x 3.3 x 10\(^5\) Heat lost = Heat gained ∴ M x 3.3 x 10\(^5\) = 2 x 400 x 100 ∴ M =\(\frac{(2 \times 400 \times 100)}{3.3 \times 10^5}\) = \(\frac{8}{33}\)Kg

The general wave equation is given by; y = A Sin(wt - 2\pix/Wavelength) where w = angular velocity Thus, comparing this equation with the equation y = 0.25 x 10-3 Sin (500t - 0.025x) That Amplitude = 0.25 x 10-3m The angular velocity(w) = 500 (the coefficient of t in both equations) ∴ w = 500 rad s\(^{-1}\) = 5.00 x 10\(^2\)rad s\(^{-1}\)

When a sound wave moves from a region of cold air to a region of hot air, its wavelength will increase. This occurs because sound travels more quickly in warmer air due to the heightened energy of the air molecules, which allows for a longer wavelength while the frequency remains unchanged.