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>
Observer A is moving inside the train
so here observer A will not be able to see the change in position of train as he is standing in the same reference frame
So here as per observer A the train will remain at rest and its not moving at all
Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body
So here observer B will see the actual motion of train which is moving in forward direction away from the platform
Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction
So the distance between them will decrease at faster rate
Now as per Newton's II law
F = ma
Now if train apply the brakes the net force on it will be opposite to its motion
So we can say
- F = ma
so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate
It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train
So there is no effect on train motion
<span>Her center of mass will rise 3.7 meters.
First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so
8.5 m / 9.8 m/s^2 = 0.867346939 s
And the distance a object under constant acceleration travels is
d = 0.5 A T^2
Substituting known values, gives
d = 0.5 9.8 m/s^2 (0.867346939 s)^2
d = 4.9 m/s^2 * 0.752290712 s^2
d = 3.68622449 m
Rounded to 2 significant figures gives 3.7 meters.
Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
Answer:
Explanation:
Let be the time required to make one revolution.
Let be the radius of the circular path.
Let be the distance travelled by ball in one revolution.
As we know,the distance travelled in one revolution is the circumference of the circle.
So,
Given,
Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.
Let be the speed of the ball.
So,the speed of the ball is
For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m
400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.