Answer:
smaller one
Explanation:
even though he is moving quicker doesn't mean he will be packing more force in the collision
First of all, we need to convert the angular speed from rev/min into rev/s:

The angular acceleration is the variation of angular speed divided by the time:

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

so, 5 revolutions.
The force result in stretching the spring 10.0 centimeters is 2.5N.
<h3>
What is Hooke's law?</h3>
If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
Learn more about hooke's law.
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Answer:9.8 m/s²
Explanation:
It was going at 9.8m/s² as this is the acceleration of an object due to gravity
when an object falls it accelerates at a consant and uniform speed which is 9.8m/s²
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