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svet-max [94.6K]
3 years ago
5

A 53 kg gymnast applies a vertical ground reaction force of 1,080 N at 0.13 m behind the center of mass during a forward somersa

ult. Assume she is facing the right. What is the torque generated about the center of mass
Physics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

Torque is 140.4 N-m.

Explanation:

Mass of the gymnast, m = 53 kg

Vertical force acting on the gymnast, F = 1080 N

Distance, r = 0.13 m behind the center of mass during a forward somersault. We need to find the torque generated about the center of mass. The force is acting behind the center of mass. F = -1080 N

Torque is given by :

\tau=F\times r\\\\\tau=-1080\times 0.13\\\\\tau=-140.4\ N-m

So, the torque generated about the center of mass is 140.4 N-m.

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Answer:

The  length is  l  =  8.6 \  m

Explanation:

From the question we are told that

   The frequencies of the two successive harmonics are f_1 =  220 \ Hz ,  f_2  =  240 \  Hz

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Generally the frequency of a given harmonic is mathematically represented as

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Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

    220  =  \frac{n v}{2l}

and  

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So

   \frac{(n + 1 ) v}{2l} - \frac{n v}{2l}  =  240-220

=>  \frac{v}{2l}  =  20

=>   l  =  8.6 \  m

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