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Tension = Force of gravity + acceleration force = mg + ma \(a = \frac{v - u}{t}\) = \(\frac{4 - 0}{8}\) = 0.5 ms\(^{-2}\) Tension = (\(10 \times 10) + 10 \times 0.5\) = 100 + 5 = 105N

LAWS OF SOLID FRICTION: Friction opposes the relative motion of two surfaces in contact. It is independent of the area of the surface of contact. It depends on the nature of the surface. It is proportional to normal reaction (R). It is independent of the relative velocity between the surfaces. Therefore, the true statements regarding sliding friction are I, II, and IV.

Water has a density of about 0.9998 g/cm³ at 0^\circC. As water is heated from 0^\circC to 4^\circC, it actually increases in density until it reaches its maximum density at approximately 4^\circC. Once the temperature exceeds 4^\circC, the density starts to decrease as the water continues to heat up towards 30^\circC. Therefore, The density will rise and fall.

The resultant emergent velocity of a charged ion moving through combined electric and magnetic fields can be derived from the condition where the electric force equals the magnetic force. This gives us the formula for the velocity v: q E = qvB v = \(\frac{E}{B}\) (q will cancel out) NOTE: When both fields are present, for the ion to move without deflection, the electric force must equal the magnetic force.

When a charged particle is accelerated across a region of p.d. V Volt, the work done is the product of the charge Q and the p.d. V i.e. Work done = Q x V But this work done is equal to the K.E. imparted to the charge given by K.E = 1/2MV2, where V is the velocity of the charge = Q x V = 1/2MV2 ∴ p.d V = (1/2MV2)/Q = 1/2 x 9.1-31 x [4 x 107]2/1.6 x 10-19 = 8 x 9.1 x 102/1.6 = 4550V = 4.55 x 103V

Regelation occurs when ice is subjected to pressure, causing it to melt at the point of contact. When the pressure is removed, the melted water can refreeze. This phenomenon is commonly observed in scenarios where ice is under stress, such as when a wire is pulled through it with weights attached.

Power input = 3KW = 3000W Voltage = 1.5KV = 1500V Current (I) = P/V = 3000/1500 = 2Amps. ∴ Power loss in the cables = I2R = 2 x 2 x 20 = 80W

When a radioactive source emits radiations that are deflected by both magnetic and electric fields, the radiations are likely to be beta particles (\beta particles). Beta particles(\beta): These are high-energy, high-speed electrons (\beta-) or positrons (\beta+). They are negatively or positively charged, respectively, and are deflected by both electric and magnetic fields due to their charge and lower mass compared to alpha particles. Gamma rays (γ): These are high-energy electromagnetic radiation and are neutral, meaning they are not deflected by electric or magnetic fields. Alpha particles (\alpha): These are positively charged and consist of two protons and two neutrons. They are not significantly deflected by electric or magnetic fields due to their relatively large mass and low speed.

The force on the conductor is maximum when the conductor lies perpendicular to the field during which the maximum force is given as; F = ILB = 0.8 x 2 x 0.5 = 0.8N

For a photoemitted electron, eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work function = 5ev - 3ev = 2ev ∴ eV = 2eV V = 2eV/e = 2V ∴ Stopping Potential = 2.0V

In thermonuclear reactions, Energy = ∆mC\(^2\) ∆m = Change in Mass C = Speed of light ∴ 5.02 x 10\(^{-27}\) - 5.01 x 10\(^{-27}\) x [3 x 10\(^8\)]\(^2\) 0.01 x 10\(^{-27}\) x 9 x 10\(^{16}\) E = 9.0 x 10\(^{-13}\)J

The velocity V = 10 + 2t\(^2\) But acceleration, a = dv/dt thus if V = 10 + 2t\(^2\) dv/dt = 4t And t = 5s then a = 4x5 = 20m/s\(^2\) ∴ a = 20m/s\(^2\)

For transverse wave on string velocity V = \(\sqrt{\frac{T}{M}}\) where T is the tension, and M is the mass per unit length ∴ for T = 12N and M = 3.0 x 10\(^{-2}\)kgm\(^{-1}\) V = \(\sqrt{\frac{12}{3.0 \times10^{-2}}}\) V = \(\sqrt{\frac{1200}{3}}\) = \(\sqrt{400}\) ∴ V = 20m/s

In general, gases conduct electricity under reduced pressure and high voltage/potential. But from Ohms Law, current is directly proportional to voltage. Thus high current presupposes high voltage.

Tan30º = \(\frac{T_1}{100}\) T\(_1\) = 100 Tan30º