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trasher [3.6K]
4 years ago
13

A basesball is dropped from 100 meters above the surface of the earth. The baseball falls to the ground. What would happen if th

is baseball was dropped from 100 meters above the surface of the moon?
Physics
2 answers:
Readme [11.4K]4 years ago
6 0

Answer:

It would fall toward the Moon's surface

Explanation:

As you already know, a ball that is thrown 100 meters above the surface of the earth, would fall on the surface of the earth, due to the gravity of our planet, which attracts any object around it to its surface (and that is what we keep them attached to the planet's floor). The same would happen if a ball were launched 100 meters above the surface of the moon, but the ball would fall on the surface of the moon, for the same reason that is the gravity of the moon. As the Moon is much smaller than the Earth, its force of gravity is also less, therefore the bodies are attracted with less force to the lunar surface, even so the ball would fall on the surface of the moon because the distance between the ball and the earth is much greater than the distance between the moon and the ball.

Serjik [45]4 years ago
4 0

Given that, A basesball is dropped from 100 meters above the surface of the earth. If the same baseball was dropped from 100 meters above the surface of the moon, it will take more time to hit the ground as compare to the ball dropped on earth. This is moon's gravity is one-sixth of that of earth. Object falling on earth possess more force of attraction. So it will reach the earth in lesser time. At moon the force of attraction is low compare to that of earth. Object will take more time to reach the surface.

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115 divided by 2 equals 57.5
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anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

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Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

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T = 0.4113772 s

6 0
3 years ago
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Answer:

'Daniela had a 5-meter head start, and Leonard caught up to her at 25 meters.'

Explanation:

hope that helps :)

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The cart's acceleration to the right after the mass is released  is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

Upward force = Downward force

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13a = 10(9.8)

13a = 98

a = 98/13

a = 7.54 m/s²

Thus, the cart's acceleration to the right after the mass is released  is determined as 7.54 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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2 years ago
At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t
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Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

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