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valkas [14]
2 years ago
8

Football player A has a mass of 210 pounds and is running at a rate of 5.0 mi/hr. He collides with player B. Player B has a mass

of 190 pounds and is running in the same direction at 3.0 mi/hr. Which of the following statements is true?
-The momentum of player A is equal to that of player B.
-The momentum of player A is greater than that of player B.
-The momentum of player A is less than that of player B.
Physics
2 answers:
Naddika [18.5K]2 years ago
6 0

The equation for momentum is p = mv where p is the omentum, m is the mass and v is the velocity. Calculating the momentum for each football player, player A will have a momentum of 1050 lb-mi/h and player B will have a momentum of 570 lb-mi/h. Therefore, momentum of player A is greater than that of player B.

IgorLugansk [536]2 years ago
3 0

Answer:

the momentum of player A is greater than player B.

Therefore, Option 2 is correct.

Step-by-Step explanation:

We have been given the velocity of the players and mass of the players.

Player A:

Velocity is 5m/hr

Mass 210 pounds

Player B:

Velocity is 3m/hr

Mass 190 pounds

We have a formula for momentum:

p=mv

p is momentum

m is mass

v is velocity

We will  find momentums for both the players:

Momentum of player A:

on substituting the values we get:

p=210\cdot 5=1050 pounds  

Momentum of player B:

on substituting the values we get:

p=190\cdot 3=570 pounds  

Hence, the momentum of player A is greater than player B.

Therefore, Option 2 is correct.




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The position function x(t) of a particle moving along an x axis is x=4.00 - 6.00t^2

a) The point at which particle stop, it's velocity = 0 m/s

  So dx/dt = 0

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  So when time t= 0, velocity = 0 m/s

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At t = 0 the particle is (momentarily) stop

b) When t = 0

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SO at x = 4m the particle is (momentarily) stop

c) We have x=4.00 - 6.00t^2

   At origin x = 0

  Substituting

         0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}

         t = 0.816 seconds or t = - 0.816 seconds

So when  t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

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2 years ago
Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
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Answer:

Total Work done =0.65 joule

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The work done to move the toy accros the first meter is

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The work done to move the toy across the next 2m at an angle of 30° is

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Answer: A. Greater than 384 Hz

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Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

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7 0
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