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matrenka [14]
3 years ago
14

What stores energy for a quick release in a cell?

Physics
1 answer:
luda_lava [24]3 years ago
5 0
It is the mitochondria of a cell that stores energy for a quick release. <span>Mitochondria break down glucose to release the energy for cells to use. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
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"The work done on an ideal gas system in an isothermal process is -400 J. What is the change in internal (thermal) energy of the
Archy [21]

Answer:

zero

Explanation:

Work done = - 400 J

In an isothermal process, the temperature remains constant. So, according to the first law of thermodynamics

dQ = dU + dW

As the temperature remains constant and the change in internal energy is the function of change in temperature, here change is temperature is zero so the change in internal energy is also zero.

6 0
3 years ago
What are the conditions of equilibrum
natta225 [31]

Answer:

they are given below

The sum or resultant of all external forces acting on the body must be equal to zero.

The sum or resultant of all external torques from external forces acting on the object must be zero.

3 0
2 years ago
Read 2 more answers
O D. Both objects won't move at all. They will just stay where they were released.
astraxan [27]

Answer:

Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.

Probably D:  Soccer ball was affected by air resistance more than the golf ball.

Explanation:

Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.

In a vacuum, (no friction) both objects fall at the same rate regardless of mass.

6 0
3 years ago
A car accelerates uniformly in a straight line
julia-pushkina [17]

The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of

<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) <em>t</em>²

<em>t</em>² = 2 (69 m) / (2.8 m/s²)

<em>t</em> ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

8 0
3 years ago
A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu
tekilochka [14]

Answer: 0.43\ rad/s

Explanation:

Given

Mass of child m=34\ kg

speed of child is v=2.8\ m/s

Moment of inertia of merry go round is I=510\ kg.m^2

radius r=2.31\ m

Conserving the angular momentum

\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s

7 0
2 years ago
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