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Romashka-Z-Leto [24]
3 years ago
10

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

ng the energy each time his feet touch down, there’s an average 10% loss with each compression. What must the runner’s additional power output be to account for just this loss, if he averages 0.8 s per stride?
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

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Answer::

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Explanation:

From the question we are told that

      The  distance of separation is  d =  0.047 \ mm =  0.047 *10^{-3} \ m

       The  distance of the screen is  D =  6.60 \ m

      The  width of the fringe is y  =  8.9 \ cm  =  0.089 \ m

     

Generally the width of the width of the fringes is mathematically represented as

          y  = \frac{\lambda  *  D }{d }

=>       \lambda  =  \frac{y *  d }{D }

=>      \lambda  =  \frac{ 0.089  *   (0.047 *10^{-3}) }{6.60  }

=>    \lambda  =  634 *10^{-9}

=>  \lambda  =  634 nm

Generally the speed of light is mathematically represented as

         c  =  f * \lambda

=>       f=  \frac{ c}{\lambda }

=>         f=  \frac{ 3.0 *10^{8}}{634 *10^{-9}}

=>     f =  4.73 *10^{14} \  Hz

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