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const2013 [10]
2 years ago
7

Two billiard balls roll towards each other. They each have a mass of 0.3 kg. ball 1 is moving at 1 m/s to the right while ball 2

is moving at 1 m/s to the left. Calculate the momentum of the system.
with explanation pls ​
Physics
1 answer:
exis [7]2 years ago
8 0

Answer:

mass= 0.05kg

v=6m/s

pi=mvi=0.05*6=0.3kgm/s

p1f=0.3kgm/s

p2f=0.3kgm/s

p1f-p1i=0.3-0.3=0.6kgm/s

p2f-p2i=0.3(0.3) =0.6kgm/s

it was in opposite directions

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at the submicroscopic and microscopic levels temperature is proportional to the average ______ kinetic energy of the particles o
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A projectile is launched at ground level with an initial speed of 50.0m/s at an angle of 30° above the horizontal. It strikes a
tino4ka555 [31]

Answer:

x = 129.9 m

y = 30.9 m

Explanation:

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

v_{i} = 50 m/s

Angle = 30°

Time = t = 3 s

horizontal component of velocity = v_{i_{x}} = v_{i}cos30°

v_{i_{x}} = 50cos30°

v_{i_{x}} = 43.3 m/s

Vertical component of velocity = v_{i_{y}} = v_{i}Sin30°

v_{i_{y}} = 50Sin30°

v_{i_{y}} = 25 m/s

This is a projectile motion, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

But the vertical component of velocity varies with time and there is an acceleration along vertical direction which is equal to gravitational acceleration g.

Horizontal distance = x =  v_{i_{x}}t

x =  43.3*3

x = 129.9 m

Vertical Distance = y = v_{i_{y}}t -0.5gt²

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3 0
3 years ago
A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ram
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Answer:

a) Maximum height reached above ground = 2.8 m

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Explanation:

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         0²=5.6² + 2 x -9.81 x s

          s = 1.60 m

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b) We have equation of motion v= u+at

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         0= 5.6 - 9.81 x t

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  We have equation of motion s =ut + 0.5 at²

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          a = 0 m/s²

          t = 0.57  

  Substituting

         s =3.5 x 0.57 + 0.5 x 0 x 0.57²

         s = 2 m

  When he reaches maximum height he is 2 m far from end of the ramp.

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