Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared to the incident ray, w
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Answer:
Approximately if that athlete jumped up at
. (Assuming that
.)
Explanation:
The momentum of an object is the product of its mass
and its velocity
. That is:
.
Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: and
. Therefore:
and
.
Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.
Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:
.
Therefore:
.
.
Rewrite this equation to find an expression for , the speed of the earth after the jump:
.
The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is .
.
Calculate using
and
values from the question:
.
The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately .
Answer:
The current is
Explanation:
From the question we are told that
The radius is
The current density is
The distance we are considering is
Generally current density is mathematically represented as
Where A is the cross-sectional area represented as
=>
=>
Now the change in current per unit length is mathematically evaluated as
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
substituting values