What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
1 answer:
The elastic potential energy of the spring is 6.8 J
Explanation:
The elastic potential energy of a compressed/stretched spring is given by the equation:
where
k is the spring constant
x is the elongation of the spring
The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):
By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,
Therefore when the spring has a elongation of , its potential energy is
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Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
Finally, you obtain for E:
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C