A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

Answers

Answer 1

Answer: IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy = (mass) x   (gravity) x (height)

   = (0.15 kg) x (9.8 m/s²) x (20 m)

   = 29.4 kg-m²/s² = 29.4 joules .

Answer 2

Answer: The conservation of energy says that energy in an isolated system remains constant, that it is not created nor destroyed but transforms from one form of energy to another.
So when the ball is thrown into the air, it has kinetic energy (KE). As it goes into the air, that kinetic energy transforms into gravitational potential energy (GPE). The higher it gets, the more GPE it has and therefore the less KE it has.
At the top, all of it is transformed into GPE and there is no KE.
But total energy at any point after the ball is thrown until it lands it has the same energy throughout.

Knowing this, you can set the beginning KE equal to the final GPE.
KE = GPE

Insert formula for GPE.

KE = mgh
KE = 0.15kg*9.8m/s^2*20.0m
KE = 29 J

The ball initially had 29 J of kinetic energy

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Answers

a ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
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Answers

Answer : Option D) Fixed-position pulleys are connected to a fixed point and do not move.

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Movable pulleys are ones which has one part of the rope attached to a fixed object, like a bar or a beam. Whereas when both the parts of a rope are attached to a fixed object, this is called as a fixed pulley.

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Answers

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Answers

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The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth exerts a gravitational force of 9.1 × 104 newtons on the telescope. The magnitude of Earth’s gravitational field strength at this location is

Answers

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Answers

Answer:

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Explanation:

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