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Bogdan [553]
3 years ago
9

An airplane is flying in a horizontal circle at a speed of 92.1 m/s. The 55.6 kg pilot does not want his radial acceleration to

exceed 8.45 g. The acceleration of gravity is 9.8 m/s². What is the minimum radius of the circular path?
Physics
1 answer:
Fudgin [204]3 years ago
7 0

Answer:

 r = 102.43 m

Explanation:

Newton's second law for this case is

              F = ma

Where the acceleration is centripetal

            a = v² / r

            r = v² / a

They indicate that the radial acceleration is 8.45 g

           r = v² / 8.45 g.

           r = 92.1² / 8.45 9.8

           r = 102.43 m

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The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa
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Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

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Answer:

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Explanation:

I'm so bored

yrfgggghhgghhyuj

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