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In a traveling wave, energy is what is propagated (transferred) from one point to another through the medium. Frequency, amplitude, and wavelength remain constant (in a uniform medium) but are not the quantity being "carried along" the wave.

in a vibrating string the fundamental frequency is given by fo = v/2L therefore velocity = 2L x f. = 2 x 0.6 x 220 = 264 m/s.

From the relation PV = nRT V = 10-33; n = 3 moles; T = 27 + 273 = 300k R = 8.3 j mol-1 K-1 => P = nRT/V = 3 X 8.3 X 300/10-3 = 7.47 X 106 Nm-2

The energy store is equal to the work done in stretching the wire from E to F given by the \(\frac{1}{2}\)Fe where e = 0.10 - 0.50 = 0.05m. F = 0.2 + 0.1 = 0.3N = \(\frac{1}{2}\) x (0.2 + 0.1) x (0.10 - 0.50) = 7.5 x 10\(^{-3}\)J

The work done on this object is given by the area of the triangle, which is equal to 1/2 fx

A photon of energy E = hf = 6.63 x 10-34 x 3 x 105 = 19.89 x 10-29 = 2.0 x 10-28J

Power is transmitted at very high voltages primarily to reduce heating in the coils(cables). High Voltage: Lowers current, minimizing resistive (I\(^2\)R) losses, which cause heating. Eddy Currents: While they can be a concern in certain applications, such as transformers, they are not the primary issue in high-voltage transmission.

The energy stored in a given capacitor is given by E = \(\frac{1}{2}\) CV\(^2\) = \(\frac{1}{2}\) QV = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\) Where C = the capacitance of the capacitor Q = charge carried by the capacitor V = p.d across the plates of the capacitor, thus from the data given, C = 10\muF = 10 x 10\(^{-6}\)F Q = 100\muc = 100 x 10\(^{-6}\)C ∴ Energy stored = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\) = \(\frac{1}{2}\)\(\frac{(100 \times 10^{-6})^2}{10 \times 10^{-6}}\) = 0.0005 = 5 x 10\(^{-4}\)J

1 KW = 1000 watts, power P = IV, thus for a mains of 250 V, the maximum permissible Current I = \(\frac{\text{P}}{\text{V}}\) = \(\frac{1000}{250}\) = 4 A Thus, the fuse rating in the plug should be 4 A Putting safety into consideration, the appropriate rating of the fuse should be 5A That is = 4 x 1.25 = 5A ( 1.25 = safety factor)

Electric Power = IV = I2R = V2/R thus for a device rated 1000 W, 230V, from the relation P = V2/R, 1000 = (230)2/R => R = (230)2/1000 = 52900/1000 = 52.9Ω

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

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

h = \(\frac{p}{eg} = \frac{9100}{2000 \times 10} = 0.455m\)

R.d = \(\frac{\text{weight in air}}{\text{apparent loss of weight in water}} = \frac{45}{45 - 15}\) = \(\frac{45}{30} = 1.50\)

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