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Elena-2011 [213]
2 years ago
6

1.) if the total momentum for a system is the same before and after the collision, we say that momentum is conserved. if momentu

m were conserved, what would be the ratio of the total momentum after the collision to the total momentum before the collision?
2.) if the total kinetic energy for a system is the same before and after the collision, we say that kinetic energy is conserved. if kinetic energy were conserved, what would be the ratio of the total kinetic energy after the collision to the total kinetic energy before the collision?
Physics
1 answer:
gogolik [260]2 years ago
6 0

From the law of the conservation of energy;  the ratio of the total kinetic energy after the collision to the total kinetic energy before the collision must be 1.

<h3>What is momentum?</h3>

The term momentum is the product of mass and velocity. The principle of conservation of linear momentum states that total momentum before collision must be the same as the total momentum after collision thus the  ratio of the total momentum after the collision to the total momentum before the collision must be 1.

Also, from the law of the conservation of energy;  the ratio of the total kinetic energy after the collision to the total kinetic energy before the collision must be 1.

Learn more about momentum:brainly.com/question/24030570

#SPJ1

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Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

T=2\pi\sqrt{\frac{I}{K}}

I=mR^{2}

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}

As dimensions of new balance wheel are one-third of their original values

R_{new}=\frac{R}{3}

\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega

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What is the difference between a distance time graph and a speed time graph?
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a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav
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The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Time to reach ground from maximum height (t) = 2.2 s
  • Initial velocity (u) = 0 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
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<h3>How to determine the velocity when the ball strikes the ground</h3>

The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

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