Answer:
The spring constant is 60,000 N
The total work done on it during the compression is 3 J
Explanation:
Given;
weight of the girl, W = 600 N
compression of the spring, x = 1 cm = 0.01 m
To determine the spring constant, we apply hook's law;
F = kx
where;
F is applied force or weight on the spring
k is the spring constant
x is the compression of the spring
k = F / x
k = 600 / 0.01
k = 60,000 N
The total work done on the spring = elastic potential energy of the spring, U;
U = ¹/₂kx²
U = ¹/₂(60000)(0.01)²
U = 3 J
Thus, the total work done on it during the compression is 3 J
If magnification is less than one it means the image is the same exact size as the object.
Any two objects in the universe attract each other. Gravity is the force exerted by earth on you (you exert the same force on earth) but due to the fact that earth has a huge mass compared to yours, you will be attracted to earth only by a small gravitational force.
If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m
