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3 years ago

If you tie a bowling ball onto the ceiling and let it swing why doesn't it hit you in the face?

1 answer:

7 0

The bowling ball starts with a certain amount of potential energy. It gets this energy because someone had to lift it up to hang it. Bowling balls are heavy and you have to do work to get it up in the air. Whenever we do work on something, we are giving it energy. When you let go of the ball, it swings downward like a pendulum. As it starts swinging, the energy changes from potential energy to kinetic, or moving, energy. The total amount of energy, moving plus stored, stays the same; it only changes form. When the ball swings back to where it started, the energy changes back to potential energy. Since the total energy has to stay constant, the kinetic energy of the ball must be zero and the ball must stop moving. It can’t hit you

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How can you safely experiment with plants during a scientific investigation?

liubo4ka [24]

I would think the answer is c.

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3 years ago

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What is the magnitude of the centripetal

zepelin [54]

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7.5

Explanation:

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The National Aeronautics and Space Administration (NASA) studies the physiological effects of large accelerations on astronauts.

m_a_m_a [10]

Answer:

$v = 23.95 m/s$

Explanation:

As we know that when astronaut is revolving in circular path then the acceleration of the astronaut is due to centripetal acceleration

so it is given as

$a_c = \frac{v^2}{R}$

here we know that

$a_c = 4.50 g$

also we know that

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4 years ago

The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m

lions [1.4K]

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

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3 years ago

How do longitudinal waves transport energy?

marissa [1.9K]

A longitudinal wave transports energy through the medium without permanently transporting matter.

Hope this helps :D

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3 years ago