The answer is 789.25 which you’d subtract 2011.25-122.2 I think sry if I’m wrong
The spring constant is 147 N/m
Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm
We need to find the spring constant
A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring
We know that F = kx
300(9.8)= k (0.02)
k = 147.15 N/m
Rounding off to the nearest is 147N/m
The spring constant is 147N/m
Learn more about Hooke's law here
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Answer:
595391.482946 m/s

Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron = 
t = Time taken = 1 second
m = Mass of proton = 
Velocity of proton is given by

The speed of the proton is 595391.482946 m/s
Current is given by

Number of protons is

The number of protons is 
If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.