Two objects of equal mass collide on a horizontal frictionless surface. Before the collision, object A is at rest while object B
1 answer:
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THere is a standard relationship that gives this result where the capacity of the capacitor is used:
.
We know though that Q/c=V and thus we can use the relationship:
E=Q*V/2 where we have just substituted in. If we also take into account that Q=VC, we can also get that E=V^2*C/2.
We are given the charge and the potential, so the best expression to use is the middle one.
Substituting, we get that E=1/2*8*10^(-10)*20=8*10^(-9).
The answer is B
We know though that Q/c=V and thus we can use the relationship:
E=Q*V/2 where we have just substituted in. If we also take into account that Q=VC, we can also get that E=V^2*C/2.
We are given the charge and the potential, so the best expression to use is the middle one.
Substituting, we get that E=1/2*8*10^(-10)*20=8*10^(-9).
The answer is B
Answer:
d =
Explanation:
This comes from diffrection pattern equation used to locate fringe pattern, i will explain every variable in it here.
n = order of fringe = 0 1 2 3 .. and negative integars as well.
d is sacing between slits.
Theta = angle at which light-Ray is directed towards fringe.
It appears that the distance between consectice fringes would same as the distance between two slits 'd'.
therefore calculating that distance should give us the distance between two fringe patters.
Note we don't use transversal seperation distance here.