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ANEK
3 months ago
8

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

con rotates at 10 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles along the shore from the point A?
Physics
1 answer:
serg [3.5K]3 months ago
6 0

Answer:

v = 3369.2 m/s

Explanation:

The beacon is rotating at an angular speed of

f = 10 rev/min

so we have

\omega = 2\pi f

\omega = 2\pi(\frac{10}{60})

\omega = 1.047 rad/s

We know that

v = r \omega

At this point we have

r = 2 miles = 2(1609 m)

r = 3218 m

So we can conclude with

v = 3218(1.047)

v = 3369.2 m/s

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