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3 years ago

Forces normal to a particle's displacement do no work.

Physics

1 answer:

snow_tiger [21]3 years ago

3 0

Answer:

Explanation:

The work done is defined as the product of force applied in the direction of displacement and the displacement.

W = F x d x Cosθ

where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.

For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.

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$\frac{d_{1}}{d_{2}}=0.36$

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Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

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$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

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