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mamaluj [8]
3 years ago
12

What is the gravitational potential energy of a 3 kg ball that is 1 meter above the floor?

Physics
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer:

PE=mgh

M= Mass (kg)

G= Gravitational field strength (N/kg)

H= Hight (m)

PE= Gravitational Potential Energy (J)

Explanation:

Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.

You might be interested in
Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of an average person i
faust18 [17]

Well first of all, you must realize that it depends on how the jumpers are distributed on the earth's surface.  If,say, one billion of them are in the eastern  hemisphere and the other billion are in the western one, then the sum of all of their momenta could easily be zero, and have no effect at all on the planet.  I'm pretty sure what you must have in mind is to consider the Earth to be a block, with a flat upper surface, and all the people jump in the same direction.

average mass per person = 60 kg.
jump velocity = 7 m/s straight up and away from the block, all in the same direction
one person's worth of momentum = (m) (v) = 420 kg.m/s
sum of two billion of them = 8.4 x 10¹¹ kg-m/s all in the same direction

Earth's "recoil" momentum = 8.4 x 10¹¹ in the opposite direction = (m) (v)

Divide each side by 'm' :     v = (momentum) / (mass) =

The Earth's "recoil" velocity is   (8.4 x 10¹¹) / (5.98 x 10²⁴) = 

                                                               1.405 x 10⁻¹³ m/s =

                                              <em> 0.00000000014 millimeter per second

</em>
I have no intuitive feeling for this kind of thing, so can't judge whether
the answer is reasonable.  But my math and physics felt OK on the
way to the solution, so that's my answer and I'm sticking to it.

4 0
3 years ago
A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
Mekhanik [1.2K]
<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0
3 years ago
A galaxy that is a featureless spherical ball of stars would be called a type
satela [25.4K]

Answer:

E0

Explanation:

Yes. The "0" indicates that it is spherical

7 0
2 years ago
How far will it go, given that the coefficient of kinetic friction is 0.10 and the push imparts an initial speed of 3.9 m/s ?
Akimi4 [234]

Answer:19.5 m

Explanation:

Given

coefficient of kinetic Friction \mu _k=0.10

Initial speed u=3.9\ m/s

Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s

maximum deceleration provided by friction is

a_{max}=\mu _k\cdot g

a_{max}=0.1\times 3.9=0.39\ m/s^2

using equation of motion

v^2-u^2=2 as

where v=final\ velocity

u=initial velocity

a=acceleration

s=displacement

(0)-(3.9)^2=2\times (-0.39)\times s

s=19.5\ m

6 0
3 years ago
HELP ASAP!!
Arada [10]
Percentage error:

1.55 – 1.53 ÷ 1.53
0.02 ÷ 1.53
.013 x 100
1.3 % error

I hope this is right.
7 0
3 years ago
Read 2 more answers
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