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Fudgin [204]
3 years ago
15

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

ght and collides with a 1.2 kg mass at rest and the two masses bounce off each other elastically. find the final velocities of the two masses after the collision.
Physics
1 answer:
Elanso [62]3 years ago
8 0

First we can say that since there is no external force on this system so momentum is always conserved.

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}

0.48= 0.6*v_{1f} + 1.2*v_{2f}

0.8  = v_{1f} + 2v_{2f}

now by the condition of elastic collision

v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6

v_{2f} = 0.533 m/s

also from above equation we have

v_{1f} = -0.267 m/s

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

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olchik [2.2K]

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a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

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However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

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The change in pressure is given by the continuity equation:

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Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

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E) the flow of energy due to a temperature difference.

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Heat can be described as the flow of energy due to a temperature difference.

Which is expressed mathematically as;

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