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Rudik [331]
3 years ago
9

Your physics TA has a far point of 0.759 m from her eyes and is able to see distant objects in focus when wearing glasses with a

refractive power of −1.35 D. Determine the distance between her glasses and eyes.
Physics
1 answer:
romanna [79]3 years ago
4 0

Answer:

d=0.019m

Explanation:

From the question we are told that:

Far point x=0.759m

Refractive power P=-1.35 D.

Generally, the equation for Focal length is mathematically given by

F=\frac{1}{P}

F=\frac{1}{-1.35}

F=-0.74m

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Where

u=o

\frac{1}{-0.74}=\frac{1}{0}+\frac{1}{v}

v=-0.74m

Therefore,The between her glasses and eyes

d=x-v

d=0.759-0.74m

d=0.019m

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Answer:

t = 0.029s

Explanation:

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The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the
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e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

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Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0
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