The force result in stretching the spring 10.0 centimeters is 2.5N.
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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:
u= 187.61 ft/s
Explanation:
Given that
g= - 32 ft/s²
The maximum height ,h= 550 ft
Lets take the initial velocity = u ft/s
We know that
v²=u² + 2 g s
v=final speed ,u=initial speed ,s=height
When the object reach at the maximum height then the final speed of the object will become zero.
That is why
u²= 2 x 32 x 550
u²= 35200
u= 187.61 ft/s
That is why the initial speed will be 187.61 ft/s
Balance and beaker of water. The balance will measure mass and the beaker will measure the volume when you take the initial volume without the key subtracting that value from the value with the key in the beaker