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aivan3 [116]
3 years ago
9

A 2-kg bowling ball sits on top of a building that is 40 meters tall.

Physics
2 answers:
Vesna [10]3 years ago
7 0
A 2-kg bowling ball sits on top of a building that is 40 meters tall possess gravitational potential energy. It <span> is </span>energy<span> an object possesses because of its position in a </span>gravitational <span>field. We calculate as follows:

GPE = mgh = 2(9.8)(40) = 784 J</span>
Dahasolnce [82]3 years ago
6 0
The bowling ball is at rest, so it only has gravitational potential energy.

Ug = mgy
Ug = (2)(9.8)(40) = 784 J

Need any more help?
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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
If a racquet ball hits the wall with 20 N of force, how much force will it bounce back with?
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Answer:

-20N

Explanation:

The racquet ball will bounce back with the same force.

This is in compliance with newton's third law of motion:

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If the ball hits the wall with an action force of 20N, the reaction force will be -20N.

The negative indicates an oppositely directed force.

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