A
B
C
D
E
Correct answer is E
No explanation has been provided for this answer.
4.9 x 10-9N
1.3 x 10-6N
6.6 x 10-8N
6.6 x 10-6N
2.6 x 10-9N
Correct answer is D
F = \(\frac{Gm_1m_2}{r^2} = \frac{6.6 \times 10^{-11} \times 80 \times 50}{0.2 \times 0.2}\)
= 6.6 x 10-6N
A force of 500N1 is applied to a steel wire of cross-sectional area 0.2m2, The tensile stress is
2.5x104Nm-2
1.0x102Nm-2
1.0x103Nm-2
2.5x103Nm-2
Correct answer is D
\(Stress = \frac{Force}{Area}\)
\(Stress = \frac{500}{0.2} = 2500\)
= 2.5 x 103Nm-3
200J
4000J
2000J
500J
Correct answer is C
Kinetic Energy = mgh
= 20 x 10 x 10 J
= 2000J
What happens when three coplanar non-parallel forces are in equilibrium?
Their lines of action are parallel.
They are represented in magnitude only
They are represented in direction only
Their lines of action meet at a point
Correct answer is D
Conditions for non-parallel coplanar forces
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