Displacement is usually how the messure the rock
The correct answer is D, Diamond
In this problem we have the electric field intensity E:
E = 6.5 × 
newtons/coulomb
We have the magnitude of the load:
q = 6.4 × 
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × 
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × 
)(6.5 × )(1.2 × )
PE = 5.0 x 
joules
None of the options shown is correct.
The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
To learn more about work done click on the given link brainly.com/question/25573309
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