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3 years ago

what is the work done by a force equivalent to the weight of 100kg that moves an object in 7 minutes?

Physics

1 answer:

Morgarella [4.7K]3 years ago

8 0

<h3>Answer:</h3>

7000 Joules

<h3>Explanation:</h3>

We are given;

The mass of an object = 100 kg

But, Force = mass × g (taking g as 10 N/kg)

Then, Force = 1000 N

Assuming we are given the distance moved by the object as 7 m

We can calculate the work done by the force

We need to know that;

Work done = Force × distance

Therefore;

Work done = 1000 N × 7 m

= 7000 Joules

Therefore, the work done by the force is 7000 Joules

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Answer:

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Answer:

From the question we are told that

The length of the rod is $L_o$

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The angle made by the rod is $\theta$

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$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

3 years ago

what is the work done by a force equivalent to the weight of 100kg that moves an object in 7 minutes?​

what is the work done by a force equivalent to the weight of 100kg that moves an object in 7 minutes?