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IRINA_888 [86]
3 years ago
6

Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. Afte

r the collision, Ball A continues to move in the same direction at 2 m/s. What is the magnitude of the velocity for Ball B after the collision?
Physics
1 answer:
timurjin [86]3 years ago
3 0

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision  = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v  = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

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antiseptic1488 [7]

Answer:

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Explanation:

The velocity of a particle(v) executing SHM is

v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

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The maximum velocity(\bf{v_{m}}) is

v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

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3 years ago
Please help me with this physics prooblem
zaharov [31]

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x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

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x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

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5 0
3 years ago
Name one contact force and one force that acts through a force field
nirvana33 [79]
A contact force is a type of force which act on an object by coming in contact with the object. Examples of contact force that acts through a force field are: applied force, frictional force, air resistance force, tension, spring force, etc.
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5 0
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Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro
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Answer:

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Explanation:

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Answer:

The acceleration is 2.2 m/s^2

Explanation:

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