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earnstyle [38]
3 years ago
5

The refracting telescope at the Yerkes Observatory has a 1.00-m diameter objective lens of focal length 20.0 m. Assume it is use

d with an eyepiece of focal length 1.90 cm. (a) Determine the magnification of Mars as seen through this telescope.
Physics
2 answers:
dangina [55]3 years ago
8 0

Magnification = (focal length of the objective) / (focal length of the eyepiece)

                     =    (20.0 m)  /  (1.9 cm)

                     =      (2,000 cm)  /  (1.9 cm)

                     =        1,053  (rounded)
igomit [66]3 years ago
7 0
Angular magnifiction=m= \frac{ f_{0} }{ f_{e} }
f_{0}==focal length of object lens=20m
f_{e}=focal length of eye lens=1.90 m
so magnification=m=20/1.9=?
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I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

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• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

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• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

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<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

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Answer:

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From the question,

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5 0
4 years ago
Read 2 more answers
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