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V125BC [204]
4 years ago
15

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

ller average density than box B. When the boxes are released, they accelerate upward to the surface. Which box has the greater acceleration?
(A) It depends on the cotent of the boxes
(B) Box B
(C) They are the same
(D) We need to know the actual densities of the boxes in order to answer the question
(E) Box A
Physics
1 answer:
sweet [91]4 years ago
4 0

Answer:

As a box A has a smaller average density than box B, then it will have a greater acceleraton towards the surface upon releasing. So the option E) Box A is a correct option.

Explanation:

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Answer:

The fraction of the total initial kinetic energy is lost during the collision is \dfrac{11}{17}\ J

Explanation:

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Mass of one piece = 300 g

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Speed of other piece = 0.75 m/s

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Using conservation of energy

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value intro the formula

300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v

v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}

v=-0.5\ m/s

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2

K.E_{i}=0.31875\ J

We need to calculate the total final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2

K.E_{f}=0.1125\ J

We need to calculate the energy lost during the collision

Using formula of energy lost

energy\ lost=\dfrac{0.31875-0.1125}{0.31875}

energy\ lost=\dfrac{11}{17}\ J

Hence, The fraction of the total initial kinetic energy is lost during the collision is \dfrac{11}{17}\ J

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