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yanalaym [24]
3 years ago
5

You pick up a 3.4 kg can of paint from the ground and lift it to a height of 1.8 m. how much work do you do on the can of paint?

you hold the can stationary for half a minute, waiting for a friend on a ladder to take it. how much work do you do during this time? you friend decides against the paint, so you lower it back to the ground. how much work do you do on the can as you lower it?
Physics
1 answer:
zepelin [54]3 years ago
5 0

1) +60.0 J

The work done when lifting an object is equal to the change in gravitational potential energy of the object:

W=\Delta U= m g \Delta h

where m is the mass of the object, g is the gravitational acceleration, \Delta h is the variation of height of the object.

In this problem, m = 3.4 kg, g = 9.8 m/s^2 and \Delta h = +1.8 m, so the work done is:

W=(3.4 kg)(9.8 m/s^2)(1.8 m)=+60.0 J

And the work done is positive, since the object has gained potential energy.


2) 0 J

In this case, there is no variation of the heigth of the object:

\Delta h =0

Therefore, the variation of potential energy is zero:

\Delta U = 0

And so, the work done is zero as well:

W=0


3) -60.0 J

In this case, the variation of height of the object is negative, since the object has been lowered to the ground:

\Delta h = -1.8 m

So, the object has lost potential energy:

W=(3.4 kg)(9.8 m/s^2)(-1.8 m)=-60.0 J

And so the work done is negative:

W=-60.0 J

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How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the o
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Question:

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes

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Answer:

S = 5.508 × 10¹¹m

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Explanation:

The radius of the orbit of Jupiter, Rj is 43.2 light-minutes

radius of the orbit of Mars, Rm is 12.6 light-minutes

Distance travelled S = (Rj - Rm)

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Answer:

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The general formula for power of a lens is:

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Therefore, for the top half of the lens:

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