3 years ago

A grapefruit falls from a tree and hits the ground 0.72 s later. How far did the grapefruit drop?

Physics

1 answer:

ycow [4]3 years ago

7 0

Answer: 2.541884 meters

Explain:

Formula: s = 0.5 * g * t^2

Time of fall (t) is 0.72 while the gravitational acceleration rate is (9.80665m/s^2).

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3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$