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3 years ago

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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olga55 [171]

Answer:

Peer Review

Explanation:

When scientist check other scientists the process is dubbed "peer review".

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3 years ago

The gravitational attraction between two objects increases if

JulijaS [17]

Answer:

they are closer!!

Explanation:

Hope this helped!! :D

8 0

3 years ago

When an object moves, stops moving, changes speed, or changes direction, how do scientists describe that condition?

lorasvet [3.4K]

Drop "moves" from the list for a moment.

You can also drop "stops moving", because that's included in "changes speed"
(from something to zero).

When an object changes speed or changes direction, that's called "acceleration".

I dropped the first one from the list, because an object can be moving,
and as long as it's speed is constant and it's moving in a straight line,
there's no acceleration.

I think you meant to say "starts moving". That's a change of speed (from zero
to something), so it's also acceleration.

8 0

3 years ago

A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma

TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

6 0

3 years ago

An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

8_murik_8 [283]

Answer:

The gplanet is 0.193 m/s^2

Explanation:

The speed of the pulse is:

$v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s$

$v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}$

where

m=mass of the wire=4 g= 4x10^-3 kg

M=mass of the object= 3 kg

Replacing values:

$g=\frac{4x10^{-3}*15.24^{2} }{3*1.6} =0.193 m/s^{2}$

7 0

3 years ago