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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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4 0
3 years ago
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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If
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Answer:  29.50 m

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f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

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x= vo*t- (a/2)* t^2

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2 years ago
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Answer:

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Given that,

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