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frez [133]
3 years ago
7

24 POINTS!!!!!!!!!!!!!!!

Physics
2 answers:
TEA [102]3 years ago
7 0
Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880

Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000

Lost Energy = 5880 - 3000 = 2880 J

In short, Your Answer would be 2880 Joules

Hope this helps!
ss7ja [257]3 years ago
6 0

Your Answer would be 2880 Joules

I took the test on K12

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The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st
irina [24]

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

4 0
3 years ago
NEED HELP NOW ETHANHUNT 25 POINTS WILL MARK BRAINLIEST!! For the independent reading all you have to do is pick a grade six book
garik1379 [7]
Can you please stop pasting this question, just go to his profile and ask him.
7 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
What factor determines an object's momentum?
vlada-n [284]

Momentum = mass x velocity

So both mass and velocity affect an object's momentum.

3 0
3 years ago
A piece of blue paper appears blue because the paper
vfiekz [6]

A piece of blue paper appears blue because the paper absorbs all colors of light except blue. <em> (b)</em>

So any light that bounces off of the paper and enters your eye must be blue light !

6 0
3 years ago
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