2015 - WAEC Physics Past Questions and Answers - page 5

E - V₁ + V₂

24V = 14V + VQP

VQP = 24 + 14V

VQP = 10V

F = \(\frac{GM_1M_2}{12^2}\)

F = 6.7 x 10^-11x

F = \(\frac{M_1M_2}{R^2}\)

F = \(\frac{6.7 \times 10^{-11} \times 6.0 \times 10^{24} \times 7.0 \times 10^{22}}{(4 \times 10^8)^2}\)

= 1.8 x 10²⁰N

E - Mec²

E = 1.2 x 10^-5 x (3 x 10⁸)²

E = 1.08 x 10¹²J

The diagram given simple illustrates that a pentagon contains three \(\Delta\)s.

Two angles which are in the ratio 2:3 will have actual values 2x^o, 3x^o respectively. Thus 2x^o + 3x^o + 3 x 60 = 3x sum of angles of \(\Delta\)a

i.e. 5x^o + 180^o = 3 x 180^o

5x^o + 180^o = 540^o

5x^o = 540^o - 180^o

5x^o = 360^o

x^o = \(\frac{360^o}{5}\)

= 72^o

Hence, the smaller of the two angles is

2 x 72^o = 144^o