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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Explanation :
There are different factors that affect the climate of an area.The distance from the equator is one of them.
Equator is closest to the sun. The rays from the sun strikes mostly near equator and gets spread over the larger area. Hence, equators are much hot as compared to that of poles.
First let us calculate for the angle of inclination using
the sin function,
sin θ = 1 m / 4 m
θ = 14.48°
Then we calculate the work done by the movers using the
formula:
W = Fnet * d
So we must calculate for the value of Fnet first. Fnet is
force due to weight minus the frictional force.
Fnet = m g sinθ – μ m g cosθ
Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48
Fnet = 84.526 N
So the work exerted is equal to:
W = 84.526 N * 4 m
<span>W = 338.10 J</span>
Eh? i don’t understand huhu