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Ede4ka [16]
3 years ago
14

While pushing a refrigerator up a ramp, there would be less friction to overcome if you used a dolly because of the wheels. So t

he efficiency of the ramp would be ________________________ with the use of the dolly.
Physics
1 answer:
FromTheMoon [43]3 years ago
4 0

While pushing a refrigerator up a ramp, there would be less friction to overcome if you used a dolly because of the wheels. So the efficiency of the ramp would be greater with the use of the dolly.

<u>Explanation:</u>

The ramp provides for a wedge setup, so when the refrigerator is being pushed up the ramp, the friction on the ramp would equally distributed to the refrigerator when pushed wholly as the friction applied is a product of the coefficient of friction with the Normal reaction of the weight in contact with the surface.

Therefore, when the dolly is used, the surface in contact becomes divided and thus, the friction applied would become less due to though coefficient of friction remains the same, the normal reaction of the weight becomes small and so does the applied friction. Therefore, efficiency of the ramp increases as the friction is less and work done is more.

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Answer:

1. The car which is twice as massive as the other will have twice potential energy.

Explanation:

The potential energy of an object is given by:

U=mgh

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h is the height of the object measured to some reference level

In this problem, we have two cars at same elevation (= same h in the formula) but one car has twice the mass of the second car. Calling m the mass of one car and 2m the mass of the second car, we have:

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U=mgh

- potential energy of the second car:

U'=(2m)gh=2mgh=2U

So, the car which is twice as massive as the other will have twice potential energy.

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4 years ago
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5.38
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4 0
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An object is placed a distance of twice the focal length away from a diverging lens. What is the magnification of the image?
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Explanation:

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So we can find q from the equation:

\frac{1}{q}=\frac{1}{(-f)}-\frac{1}{p}=-\frac{1}{f}-\frac{1}{2f}=-\frac{3}{2f}\\q=-\frac{2}{3}f

Now we can find the magnification of the image, given by:

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