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3 years ago

5

A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer

's velocity at the edge of the roof if it is moving at a speed of 8.25 m/s

1 answer:

dlinn [17]3 years ago

7 0

<h2>Answer:7.14 $ms^{-1}$ ,4.125 $ms^{-1}$ </h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let $v$ be the total velocity.

Given that, $v=8.25ms^{-1}$

We call the horizontal velocity as $v_{h}$ and the vertical velocity as $v_{v}$ .

$v_{h}$ = $vCos\alpha$

$v_{v}=vSin\alpha$

where $\alpha$ is the angle between the object and horizontal.

It is given that $\alpha =30^{0}$

$v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}$

$v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}$

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From the question we are told that

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Now making $v_n$ the subject

$v_n = \sqrt{\frac{2}{\rho} [ P_i - Pn + \frac{1}{2} \rho v_i^2}$

substituting values

$v_n = \sqrt{\frac{2}{1000} [ 200000 - 101300 + \frac{1}{2} (1000 * (1.0)^2)}$

$v_n =14.09 \ m/s$

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3 years ago