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yarga [219]
3 years ago
7

A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W

hen a compressed spring attached to one of the carts is released, the carts are pushed apart. After the spring is released, the speed of cart B is 6.0 meters per second, as represented in the diagram belowWhat is the speed of cart A after the spring is released?
Physics
1 answer:
7nadin3 [17]3 years ago
7 0
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
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In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
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Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

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     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

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    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

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   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

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Now  since capacitance is  constant  at  \tau  =  3

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