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marysya [2.9K]
3 years ago
7

A perfect bouncy ball would continue bouncing the same height forever. But in reality, the ball stops bouncing because energy is

"lost" to ...
A.sound energy
B.compression of the ball
C.heat energy absorbed by the ball
D.all answers are true
Physics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

D. all answers are true

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The relationship that exists between gravity and distance and mass respectively.
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Answer:

D...............................

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I WILL MARK YOU THE BRAINLIEST NO LINKS<br>What goes were put them in the correct place
PIT_PIT [208]

Answer:

Fluid fricton goes to Static friction and sliding friction goes to rolling friction

Explanation:

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2 years ago
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A drag racing vehicle travels from zero to a hundred miles per hour and 5 seconds north what is acceleration
avanturin [10]
Acceleration(a) is the change in velocity(Δv) over the change in time(Δt). so just divide your velocity and time.
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2 years ago
suppose you have a 69.0-kg wooden crate resting on a wood floor. what maximum force can you exert horizontally on the crate with
marusya05 [52]

You've got a 69.0-kg wooden crate on a wooden floor. The box can withstand a force of up to 338N in a horizontal direction without being moved. Following this, the wooden creates moving stats.

In order to calculate the friction coefficient, divide the force pushing two objects together by the force acting between them. friction coefficient might be 0 or one. They can be split into two categories: friction coefficient that is static. Kinetic friction coefficient (also known as sliding coefficient of friction).

the acceleration brought on by the gravitational pull of large masses generally, gravitational , often known as the acceleration brought on by the Earth's gravitational pull and centrifugal force,

F= friction coefficient *M*g

F= 0.5*69*9.8

F=338N

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7 0
1 year ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
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