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V\(_{max}\) = \omegar \omega = 2\(\pi\)f but f = \(\frac{1}{T}\) \omega = \(\frac{2\pi}{T}\) = \(\frac{2\pi}{4\pi}\) ( since T = 4\(\pi\) given) \omega = 0.5rads/secs V\(_{max}\) = \omegar = 0.5 x 3 = 1.5m/s ( r = 3cm given)

The total momentum of the two balls after collision: Can be determined. The total momentum before the collision can be calculated using the mass and velocity of the moving ball, and since momentum is conserved, the total momentum after the collision will be equal to the total momentum before the collision

Work done = area under the triangle Work done = \(\frac{1}{2}\) x base x height = 0.5 x 20 x 100 = 1000J Therefore, the work done, as calculated from the force-displacement diagram, is 1000J

Efficiency = \(\frac{ poweroutput}{powerinput}\) but power output = \(\frac{ f \times s}{ t}\) = \(\frac{75 \times 10 \times 3}{ 30}\) = 75W Efficiency = 80%, power output = 75w then 0.8 = \(\frac{ 75}{powerinput}\) power input = \(\frac{ 75}{0.8}\) = 93.75W.

When two bodies are in thermal equilibrium, it means that they have the same temperature. This is because, in thermal equilibrium, there is no net transfer of heat between the bodies, indicating that they have reached the same temperature.

γ\(_r\) = γ + γ\(_{app}\) 1.8 x 10\(^{-4}\) = (3 x 8.0 x 10\(^{-6}\)) + γ\(_{app}\) γ\(_{app}\) = 1.8 x 10\(^{-4}\) - 2.4 x 10\(^{-5}\) = 1.56 x 10\(^{-4}\)K\(^{-1}\)

The point X is the Absolute zero temperature( the lowest possible temperature). Absolute zero is at -273.15 Celsius or -459.67 Fahrenheit or 0 k.

Using the general gas equation, \(\frac{ P_1 V_1}{ T_1} = \frac{ P_2 V_2}{ T_2}\) \(V_1 = 76cm^3\), \(T_1 = 27^0C\) = 27 + 273K = 300K, \(P_1\) = 80cm, at STP \(T_2 = 0^0C\) = 273K, \(P_2\) = 76cm, \(V_2\) = ? \(V_2 = \frac{ P_1\times V_1\times T_2}{P_2 \times T_1}\) \(V_2 = \frac{ 80 \times 76 \times 273}{76 \times 300}\) \(V_2 = 72.8cm^3\)

mass of metal = 50g = 0.05kg, c = 45Jkg\(^{-1}\)k\(^{-1}\), \DeltaT = 80 - 20 = 60ºC Quantity of heat lost(Q) = mass of metal x specific capacity of mass x change in temperature Q = m x c x \DeltaT Q = 0.05 x 45 x 60 = 1350J = 1.35 x 10\(^3\)J

Relative humidity = \(\frac{\text{mass of water vapour in a given volume of air(m)}}{\text{mass of water vapour needed to saturate same volume of air at same temp(M)}}\) x 100% R.H = \(\frac{\text{m}}{\text{M}}\) = \(\frac{0.05}{0.15}\) x 100% Therefore, relative humidity of the air = 33.33%

The temperature at which the saturated vapour pressure of a liquid is equal to the external atmospheric pressure is known as its BOILING POINT.

Given: distance(d) = 20.0m. time(t) = 1secs, f = 2 x 10\(^3\)Hz, wavelength(\lambda) = ? But v = \(\frac{s}{t}\) =20.0m/s = f\lambda \lambda = \(\frac{\text{v}}{\text{\lambda}}\) = \(\frac{20}{2\times10^3}\) = 0.01m = 1.0 x 10\(^{-2}\)m

y = A sin(\omegat + Φ ) A = 4cm, Φ = \(\frac{2}{3} \pi \) rad, \omega = \(\frac{1}{3}\pi\ rads^{-1}\) substituting the given values y = 4 sin ( \(\frac{1}{3}\pi\)t + \(\frac{2}{3} \pi \)) y = 4sin \(\frac{\pi}{3}\)( t + 2)

The reflected ray will rotate through angle 2 x 20º = 40º The reflected ray will now make angle 30º + 30º + 40º = 100º with the incident ray(30º + 30º since i = r)

When a light source is placed at the focus of a concave mirror, the rays of light that reflect off the mirror will emerge as a parallel beam of light. This happens because when a point source of light is placed at the focal point of a concave mirror, the reflected rays leave the mirror parallel to each other.