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Sauron [17]
3 years ago
7

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

udied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm .What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Physics
1 answer:
irina [24]3 years ago
4 0

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, 31 {\rm {N}/{mm}}. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

\triangle E = 12.79 J

Explanation:

Sprinters' tendons stretch, x_s = 43 mm = 0.043 m

Non athletes' stretch, x_n = 32 mm = 0.032 m

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, E_s = 0.5kx_s^2

Maximum energy stored in the non athletes, E_m = 0.5kx_n^2

Difference in maximum stored energy between the sprinters and the non-athlethes:

\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J

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Answer:

A) Object A is 3.25 times hotter.

B) Object A radiates 111.6 times more energy per unit of area.

Explanation:

Wiens's law states that there is an inverse relationship between the wavelength in which there is a peak in the emission of a black body and its temperature, mathematically,

\lambda_{peak}= \dfrac{0.0028976}{T},

where T is the temperature in kelvins and, \lambda_{peak} is the wavelenght (in meters) where the emission is in its peak.

From here, if we solve Wien's law for the temperature we get

T=\dfrac{0.0028976}{\lambda_{peak}}.

Now, we can easily compute the temperatures.

For object A:

T_{A}=\dfrac{0.0028976}{200*10^{-9}}

T_{A}=14488K.

For object B:

T_{B}=\dfrac{0.0028976}{650*10^{-9}}

T_{B}=4458K

From this, we get that

T_{A}/T_{B}=3.25,

which means that object A is 3.25 times hotter.

Stefan's Law states that a black body emits thermal radiation with power proportional to the fourth power of its temperature.

This is

E=\sigma T^{4},

where  \sigma=5.67*10^{-8}\ Wm^{-2}K^{-4} is call the Stefan-Boltzmann constant.

From this, power can be easily compute:

E_{A}=(5.67*10^{-8}*(14488)^{4})=2.5*10^{9}W\\E_{B}=(5.67*10^{-8}*(4458)^{4})=22.4*10^{6}}W,

and we can notice that

E_{A}/E_{B}=111.6,

which means that object A radiates 111.6 time more energy per unit of area.

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3 years ago
explain why the different thermal conductivities of metal and plastic are important in the design of the bowl.
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Answer:

metal conducts heat faster than wood. which means the metal conducted heat away from your hand (or whatever it is) faster than wood.

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2 years ago
What is the linear speed of a point on the equator, due to the earth's rotation?
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ω = (2π rad)/(24*3600 s) = 7.2722 x 10⁻⁵ rad/s

The tangential velocity (linear velocity) at a point on the equator is
v = rω
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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
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Answer:

vcyl / vsph = 1.05

Explanation:

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  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
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