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dimulka [17.4K]
2 years ago
11

Two thin lenses with a focal length of magnitude 17.0 cm, the first diverging and the second converging, are located 12.8 cm apa

rt. An object 3.40 mm tall is placed 28.3 cm to the left of the first (diverging) lens. How far from this first lens is the final image formed?
Physics
1 answer:
GaryK [48]2 years ago
4 0

Answer:

1.93 cm

Explanation:

The Image Equation for the first lens will be

1/f1 = 1/do - 1/d1

Where f1 = 17cm and d1 = 28.3cm

1/do = 1/f1 + 1/d1

= 1/17 + 1/ 28.3

do = 10.62cm

This image (which is real, inverted, and enlarged) becomes the object for the second lens as we again apply the Image Equation for the second lens

1/ f2 = 1/d2 + 1/d

Since the lenses are 12.8 cm apart, the image formed 10.62 cm to the left of lens #1 is located 2.18 cm to the right of lens #2 so the object distance, then, is

1/d2 = 1/f + 1/d

= 1/17 + 1/2.18

d2 = 1.93cm

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The lowest point in Death Valley is 85 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m.What is th
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\Delta E=2.87\times 10^6\ J

Explanation:

It is given that,

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\Delta E=mg(h_f-h_i)

\Delta E=65\times 9.8(4420-(-85))

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8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
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Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

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let the distance moved by the object = d₂

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M.A = V.R

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Therefore, the distance that the object is raised above its initial position is 5.625 m.​

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