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umka2103 [35]
3 years ago
10

A hammer falls off the top roof and strikes the ground with a certain kinetic energy. If it fell from a roof twice as tall how w

ould the kinetic energy compare?
Physics
2 answers:
Akimi4 [234]3 years ago
8 0

The kinetic energy with which the hammer strikes the ground
is exactly the potential energy it had at the height from which it fell. 

Potential energy is (mass) x (gravity) x (height) .... directly proportional
to height.

Starting from double the height, it starts with double the potential
energy, and it reaches the bottom with double the kinetic energy.

SOVA2 [1]3 years ago
3 0
No it would not compare because if you dropped off a higher building the kinetic energy would be higher than it was before.
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Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars. (T/F)
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Answer:

Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars.

False

Explanation:

The “hot Jupiters” joint word startes to be used to be able to describe planets like 51 Pegasi b, a planet with a 10-day-or-less orbit and a mass 25% or greater than Jupitere, circling a sun-like star planet in 1995, which was found by astronomers Michel Mayor and Didier Queloz, who were awarded the 2019 Nobel Prize for Physics along with the cosmologist James Peebles for their “contributions to our understanding of the evolution of the universe and Earth’s place in the cosmos.”

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After a wave passes through a medium, particles in the medium
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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

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PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

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\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

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a = (0.295)(0.2)

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Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

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x = 14.397*(2\pi*\frac{0.25}{2})

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