2013 - JAMB Physics Past Questions and Answers - page 2

M.A. = (\frac{Eff \times V.R.}{100})

M.A. = (\frac{90 \times 2}{100})

M.A. = 1.8

E = (\frac{L}{M.A.})

E = (\frac{180}{1.8})

E = 100N

W = (\frac{1}{2}Fe)

W = (\frac{1}{2}\times 20 \times 0.05)

W = 0.5J

H = (\frac{P}{e_g})

H = (\frac{1.01 \times 10^5}{1013 \times 10})

H = 9.97

M_p = e_p x V_p

M_p = 7.2 x 10² x 4.2 = 3024

V_A = (\frac{M}{e_A})

V_A = (\frac{3024}{8.4 \times 10^2})

V_A = 3.6

At 100^oC = 400

Therefore,

At 20^oC = (\frac{400}{100} \times \frac{20}{1}) = 80

(L_2 = L_1 + \alpha(\theta_2 - \theta_1))

= 100 + 0.00002(-10-30) = 99.92

W = (\frac{W_1 \times V_1}{V_2})

W = (\frac{10 \times 0.6}{1.2})

W = 6.0 x 10 ⁴J

MC(x \theta) = MC(y \theta)

2C = C

2 : 1

Water impurities (such as salt water) boils at higher temperature than pure water. This helps to cook the food quicker.