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Sauron [17]
4 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]4 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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vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction,\rm \mu_k=0.520

Coefficient of static friction,\rm  \mu_s=0.520

The friction force, F=?

Resolve the force in the inclined plane;

\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times  cos 25 ^0 \\\\ F= 46.2 \ N

Hence, the frictional force while the mass is sliding will be 46.2 N.

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2 years ago
Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to
Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength=\frac{velocity of sound }{frequency}

=\frac{343}{78}=4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

x-\left ( 8.2-x\right )=4.397    for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

x-\left ( 8.2-x\right )=0

2x=8.2

x=4.1m

Third

\left ( 8.2-x\right )-x=4.397

x=1.9015 m

6 0
3 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

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6 0
3 years ago
An automobile moves with a tangential speed of 50.0m/s around a circular track with a 3200.0m diameter. What is the magnitude of
DENIUS [597]

Answer: 0.031rad/s

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Tangential speed = 50.0m/s

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Radius = diameter/2

= 3200/2 = 1600m

Angular speed = tangential speed / radius

= 50/1600

= 0.031rad/s

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3 years ago
A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistan
Lubov Fominskaja [6]

The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.

<h3>What is the height of the pole vaulter?</h3>

The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.

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h = (v - u)²/2g

h = (10 - 1.1)²/2 * 9.8

h = 4.04 m.

In conclusion, the height is determined from the potential energy at that height.

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