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Energy stored in a coil of inductance, L, and carry a current, I is given by E = 1/2 L I2 = 1/2 x 5 x 10-3 x 6 x 6 = 9.0 x 10-2J

It has been scientifically proven that 100g of grapefruit contains 135g of potassium electrolyte.

Energy stored = Ivt = current x back e.m.f x time = 2 x 3 x 0.4 = 2.4J

Current (I) = q/t → q = i x t = 16 x 10-3 x 1 = 1.6 x 10-2C But 1.6 x 10-19C = I electronic charge 1 coulomb = 1 / (1.6x10-19) ∴ 1.6 x 10-2 coulombs = 1 / (1.6x10-19) x 1.6x10-2 / 1 = 1.0 x 10-2 x 1019 = 1.0 x 1017

1/C = 1/0.3 + 1/0.5 + 1/0.2 = 10.3333 => C = 1 / 10.3333 = 0.09677 ≅ 0.1μF ∴ C = 0.1μF

Pressure developed in the effort piston = F/a ∴F/a = 40/0.4 = 400/4 = 100Nm2 this pressure is transmitted to the larger piston of area A; => P = F/A ∴ 100 = 400/A => A = 400/100 = 4m2

Coefficient of friction μ = F/R = F/mg = 500/1000 = 0.5 ∴ μ = 0.5

With an increase in pressure, the viscosity of liquids except water increases while that of gases is practically independent of pressure.

Both X-rays and gamma rays are uncharged electromagnetic waves with similar high energy, but they differ mainly in their origin.

In order to protect a material from the influence of an external magnetic field, the material is placed in a soft iron ring, usually referred to as magnetic screening.

Power, P = I2R; and I = E/R But effective Parallel resistance: is given by: 1/R = 1/6 + 1/3 = 1+2/6 = 3/6 => R = 6/3 = 2Ω Current I = E/R = 12/2 = 6A But I1 = (R1/R1+R2) x 1 = 3/9 x 6 = 2A again I1 = (R1/R1+R2) x 1 = 6/9 x 6 = 4A ∴P1 = I12 R1 = 2 x 2 x 6 = 24W P2 = I22 R2 = 4 x 4 x 3 = 48W thus P1/P2 = 24/48 = 1/2 ; OR P1 : P2 = 1:2

If the angle between two vectors P and Q is 0^\circ, they point in the same direction and are parallel (in the same sense).

K.E = 1/2MV2 Force, F = ma => a = F/m = 12/4 = 3ms-2 But a = (V-U) / t => V = u + at If the initial velocity U = 0, then V = at = 3 x 3 = 9ms-2 ∴ K.E, = 1/2ms-2 1/2 4 x 9 x 9 = 162J

Component of 8N along the horizontal = 8 cos 60o = 8 x 0.5 = 4N Similarity Component of 6N along the horizontal = 6 cos 60o = 3N ∴ Total force towards the right = 10 + 4 + 3 = 17N Again total force toward the left = 4N ∴ Net horizontal force (towards the right) = 17-4 = 13N

g\(_e\) = \(\frac{GM_e}{R_e^2}\) g\(_m\) = \(\frac{GM_m}{R_m^2}\) But g\(_e\) = 80 g\(_m\) R\(_e\) = 4 R\(_m\) Divide through by g\(_m\) by g\(_e\) \(\frac{g_m}{g_e}\) = \(\frac{GM_m}{R_m^2}\) \(\div\) \(\frac{GM_e}{R_e^2}\) \(\frac{g_m}{g_e}\) = \(\frac{GM_m}{R_m^2}\) x \(\frac{R_e^2}{GM_e}\) \(\frac{g_m}{g_e}\) = \(\frac{M_m}{M_e}\) x \(\frac{R_e^2}{R_m^2}\) \(\frac{g_m}{g_e}\) = \(\frac{1}{80}\) and \(\frac{R_e}{R_m}\) = 4 \(\frac{1}{80}\) = \(\frac{M_m}{M_e}\) x 4\(^2\) \(\frac{M_m}{GM_e}\) = \(\frac{1}{80}\) x \(\frac{1}{16}\) \(\frac{M_m}{GM_e}\) = 1: 1280