What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per
1 answer:
Answer:
Explanation:
The equation for centripetal acceleration is .
We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.
Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference and the period T, then we have:
Putting all together:
You might be interested in
Answer:
Among those two medium, light would travel faster in the one with a reflection angle of (when light enters from the air.)
Explanation:
Let denote the speed of light in the first medium. Let
denote the speed of light in the air. Assume that the light entered the boundary at an angle of
to the normal and exited with an angle of
. By Snell's Law, the sine of
and
would be proportional to the speed of light in the corresponding medium. In other words:
.
When light enters a boundary at the critical angle , total internal reflection would happen. It would appear as if the angle of refraction is now
. (in this case,
.)
Substitute this value into the Snell's Law equation:
.
Rearrange to obtain an expression for the speed of light in the first medium:
.
The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.
For ,
is monotonically increasing with respect to
. In other words, for
in that range, the value of
increases as the value of
increases.
Therefore, compared to the medium in this question with , the medium with the larger critical angle
would have a larger
. such that light would travel faster in that medium.