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lesya692 [45]
3 years ago
12

At what height is an object that weighs 490 newtons if its gravitational potential energy is 4900 J?

Physics
1 answer:
Sergio [31]3 years ago
5 0

Answer:

The height at which the object is moved is 10 meters.

Explanation:

Given that,

Force acting on the object, W = F = 490 N

The gravitational potential energy, P = 4900 J

We need to find the height at which the object is moved. We know that the gravitational potential energy is possessed due to its position. It is given by :

P=mgh\\\\h=\dfrac{P}{mg}\\\\h=\dfrac{4900\ J}{490\ N}\\\\h=10\ m

So, the height at which the object is moved is 10 meters. Hence, this is the required solution.

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In all of these cases force is being applied and the object on which the force is being applied move in a horizontal or Vertical direction.  Thus Work is done !

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After a while, the car started to go around a long bend, still maintaining its constant speed of 55 miles per hour. Is there a n
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Answer:

4) True. The change of direction needs an unbalanced force

Explanation:

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3 years ago
You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
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Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

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A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

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\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

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Rationalizing:

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Let v2 be the final speed of the package dropped from a height 4H. Thus:

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Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
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