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.
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The crate would slide forward
Average speed (v) = total distance(d) / total time (t)
d = 57 + 87km = 144km
t = 2 + 3hrs = 5hrs
v = d÷t = 144÷5 = 28.8km/hr
Answer:
Explanation:
The x- and y- components of the velocity vector can be written as following:
Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:
