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WAEC Physics 2017 Paper

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Question : 17

Total: 50

A plane is inclined at an angle

\theta

to the horizontal. Its velocity ratio is

$\sin\theta$
$\tan\theta$
$\frac{1}{\sin\theta}$
$\frac{1}{\tan\theta}$

Solution:

Velocity ratio = $\frac{\text{Effort distance}}{\text{Load distance}}$

On an inclined plane, it is the inverse of $\sin\theta$ = $\frac{1}{\sin\theta}$

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