The work done by tension force of 14N applied on the laptop by a rope as it moves 2.0 mm up the slope is 0.028 J
W = F d cos θ
W = Work done
F = Force
d = Displacement
θ = Angle between force and displacement vector
F = 14 N
d = 2 mm = 0.002 m
θ = 0
W = 14 * 0.002 * 1
W = 0.028 J
Work done is the change in energy of an object. So if an object moves a certain distance, work is done on the object. If the force and displacement are perpendicular to each other there is no work done on the object.
Therefore, the work done by tension on the laptop is 0.028 J
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Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
The answer is D. Products are formed from reactants by the breaking and forming of new bonds.
Longitudinal waves transfer energy parallel to the direction of the wave motion
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