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GrogVix [38]
3 years ago
5

What is the potential energy of a 3kg ball that is on the ground?

Physics
1 answer:
ELEN [110]3 years ago
5 0

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
'relative to what?'.

         Potential energy = (mass) x (gravity) x (<em>height</em>).

So you have to specify <em><u>height above what</u></em> .

-- With respect to the ground, the ball has zero potential energy.
(If you let go of it, it will gain zero kinetic energy as it falls to
the ground.)

-- With respect to the floor in your basement, the potential energy is

                 (3) x (9.8) x (3 meters) = 88.2 joules.

(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
to the floor of your basement.)

-- With respect to the top of that 10-meter hill over there, the potential
energy is
                    (3) x (9.8) x (-10) = -294 joules

(Its potential energy is negative. After you let go of it, you have to give it
294 joules of energy that it doesn't have now, in order to lift it to the top of
the hill <em>where it will have zero</em> potential energy.)


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A ball of mass is released from rest at a height of 30 how fast is it going when it hits the ground
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Answer:

If the height is in metres, the speed is 24.25m/s

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If the moon were 2 times closer to earth than it is now, the gravitational force between earth and the moon would be
il63 [147K]
So much brighter and the moon would be so much darker than it is now because the moon is further away from the moon than it is now...
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4 years ago
You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole
Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as F_2 and the Tension in fence post as F_1, then

\sum F = 24(9.8)

F_1 + F_2= 24(9.8)

We apply summatory of moments then

F_2*1.25 = F_1*1.6

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

1.6 m - 0.35m=1.25m

We have two equation and two incognites, then replacing (1) in (2)

1.6(235.2 -F_2) = 1.25F_2

376.32 = F2(1.6+1.25)

F_2= \frac{376.32}{2.85}

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5 0
3 years ago
Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is
zepelin [54]

Answer:

mass of ball 1=m1

mass of ball 2=m2

velocity of ball=r1w1

velocity of ball 2=r2w2

Total angular momentum=m1*v1+m2*v2

but

v1=r1*w1

v2=r2*w2

Substitute values in above equation

Total angular momentum of the system=m1*r1*w1+m2*r2*w2

7 0
3 years ago
Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh
iragen [17]

For  a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity  is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

6 0
2 years ago
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