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Murljashka [212]
3 years ago
14

An object is held 24.8 cm from a lens of focal length 16.0 cm. What is the magnification of the image?

Physics
1 answer:
Gwar [14]3 years ago
3 0

Answer: 1.8

Explanation:

You are given

the object distance U = 24.8 cm

Focal length F = 16.0 cm

First find the image distance by using the formula:

1/f = 1/u + 1/v

Where V = image distance

Substitute u and f into the formula

1/16 = 1/24.8 + 1/v

1/ v = 1/16 - 1/24.8

1/v = 0.0625 - 0.04032258

1/v = 0.022177

Reciprocate both sides by dividing both sides by one

V = 45.09 cm

Magnification M is the ratio of image distance to the object distance. That is,

M = V/U

Substitute V and U into the formula

M = 45.09/24.8

M = 1.818

Magnification of the image is therefore equal to 1.8 approximately

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Answer:

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Explanation:

Let m and a denote the mass and acceleration of Spiderman, respectively.

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The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of F(\text{tension}) (which points upwards) should be greater than that of W (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

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On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

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m \cdot a = (\text{Net Force}) = F(\text{tension}) - W.

Therefore:

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By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately 8.4\times 10^{2}\; \rm N (downwards.)

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