Answer:
he correct answer is V = ER
Explanation:
In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related
ΔV = ∫ E.ds
where E is the elective field and normal displacement vector.
Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.
ΔV = ∫ E ds
ΔV = E s
since s is in the direction of the radii its value on the surface of the spheres s = R
ΔV = E R
checking the correct answer is V = ER
Motion Energy
I am writing this so it can be more than 20 letters
When geophysicists measure the geothermal gradient in areas on the ridges where there is no activity hydrothermally, the gradient is far below than what is predicted theoretically, but when measured near hydrothermal vents it is more than what is predicted. This is because most of the heat is being carried through convection by hydrothermal systems so that the average gradient when measured far from the circulation would be depressed or lower.
A pendulum is not a wave.
-- A pendulum doesn't have a 'wavelength'.
-- There's no way to define how many of its "waves" pass a point
every second.
-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.
-- The frequency of a pendulum depends only on the length
of the string from which it hangs.
If you take the given information and try to apply wave motion to it:
Wave speed = (wavelength) x (frequency)
Frequency = (speed) / (wavelength) ,
you would end up with
Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz
Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ? That's pretty absurd.
This math is not applicable to the pendulum.
Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
So overall momentum its 32kgm/s
Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball
