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GaryK [48]
3 years ago
13

What is the maximum velocity that a 0.3–kilogram mass attached to a 0.75–meter string can have if the mass is whirled around in

a circular horizontal path? The maximum tension that the string can withstand is 250 newtons.
375 meters/second

22.4 meters/second

19.4 meters/second

25 meters/second
Physics
1 answer:
amm18123 years ago
8 0
f \: = \: \frac{m {v}^{2} }{r} \\ {v}^{2} \: = \: \frac{fr}{m} \\ v \: = \: \sqrt{ \frac{fr}{m} }
f = 250 N
r = 0.75 m
m = 0.3 kg
v \: = \: \sqrt{ \frac{250 \: \times \: 0.75}{0.3} } \: \frac{m}{s}
v = 25 m/s.
Hence, Maximum Velocity = v = 25 m/s
D) 25 m/s.
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<h2>Answer :</h2>

□ The Final velocity of body = 45 m/s\sf{ }

<h2>Given :</h2>

Initial velocity of body = 0 m/s

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<h2>To Calculate :</h2>

Final velocity of body = ?

<h2>How To Calculate ?</h2>

<h3>❒ </h3>

Here in this question we are provided that a body is in rest so it's initial velocity will be 0 m/s (because from rest), Acceleration of body is 3 m/s², and Time taken is also given which is 15 sec.

<h3>❒</h3>

 To calculate the Final velocity of body we'll use the first equation of motion which says v = u + at

Where,

  • v is the Final velocity in m/s.

  • u is the Initial velocity in m/s.

  • a is the Acceleration in m/s².

  • t is the Time taken in second.

<h2>Calculation :</h2>

From first equation of motion

  • ⇒ v = u + at

  • ⇒ v = 0 + 3 × 15

  • ⇒ v = 0 + 45

  • ⇒ v = 45

Hence, the Final velocity of body is 45 m/s.

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