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:
Convection currents are the result of different heating. Lighter material (warm) rises while heavier (cold) material sinks. This movement of the materials is what causes convection currents! (BTW, it happens in water, in the atmosphere, and in the mantle of Earth!
Explanation:
I hope this helps a little! :)
Here,
height at failure, h1 = 525 m,
upward acceleration, a = 2.25 m/s^2,
velocity = v m/s,
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2,
<span>
SO, </span>
<span>
heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence,
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters
b)
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec
---------------------------------
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!</span>
Answer:
a force
Explanation:
Because if we apply force then only an object can slow down, speed up or change direction
Answer:
Explanation:
A mass of 700 kg will exert a force of
700 x 9.8
= 6860 N.
Amount of compression x = 4 cm
= 4 x 10⁻² m
Force constant K = force of compression / compression
= 6860 / 4 x 10⁻²
= 1715 x 10² Nm⁻¹.
Let us take compression of r at any moment
Restoring force by spring
= k r
Force required to compress = kr
Let it is compressed by small length dr during which force will remain constant.
Work done
dW = Force x displacement
= -kr -dr
= kr dr
Work done to compress by length d
for it r ranges from 0 to -d
Integrating on both sides
W = 
= [ kr²/2]₀^-4
= 1/2 kX16X10⁻⁴
= .5 x 1715 x 10² x 16 x 10⁻⁴
= 137.20 J
