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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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A 4 kW vacuum cleaner is powered by an electric motor whose efficiency is 90%. (Note that the electric motor delivers 4 W of net
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Answer:3.6\ kW

Explanation:

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" The work done on the car by the force of friction is 24,000 J.

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b. Completed work = force x distance. Friction force = 240 N, that is,

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<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here:  brainly.com/question/25573309

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