Answer:
8.80 Hz
Explanation:
The frequency of a loaded spring is given by
where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.
Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.
On the other hand, the frequency of the simple pendulum is affected because it is given by
where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by
It’s c hope this helps :)
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Answer:
5.5 x 10^5 N/C
Explanation:
t = 0.001 s
Δp = - 8.8 x 10^-17 kg m /s
Force is equal to the rate of change of momentum.
F = Δp / Δt
F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N
q = 1.6 x 10^-19 C
Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)
E = 5.5 x 10^5 N/C
