The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.
Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.
r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6
v - 232.5 m/s
Answer:
I am fairly certain it is A but i would wait for an expert to answer
The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.
Answer:
2 m/s²
Explanation:
Given:
v₀ = 0 m/s
v = 20 m/s
t = 10 s
Find: a
a = (v − v₀) / t
a = (20 m/s − 0 m/s) / 10 s
a = 2 m/s²
