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Lana71 [14]
3 years ago
7

A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in t

he tread of the tire (on its outer edge).
Physics
1 answer:
arsen [322]3 years ago
8 0

Answer:

v = 10.47 m/s

a = 219.32 m/s²

Explanation:

The velocity of the stone can be given by the following formula:

v = r ω

where,

v = linear velocity of the stone = ?

r = radius of tire = 0.5 m

ω = angular velocity = (200 rev/min)(2π rad/1 rev)(1 min/60 s) = 20.94 rad/s

Therefore,

v = (0.5 m)(20.94 rad/s)

<u>v = 10.47 m/s</u>

<u></u>

The acceleration of the stone will be equal to the centripetal acceleration. Therefore,

a = v²/r

a = (10.47 m/s)²/0.5 m

<u>a = 219.32 m/s²</u>

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Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is
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Complete question is;

Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at

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Explanation:

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