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3 years ago

8

During a collision, a bus runs into a skateboarder inelastically. If the mass of the skateboarders is 70 kg and is moving at 12

m/s to the right, what is the final velocity of the skateboarder if the 2000 kg bus is traveling at 25 m/s to the left before the collision?

1 answer:

agasfer [191]3 years ago

5 0

The final velocity of the skateboarder is 23.75 m/s.

We can use conservation of momentum to calculate the magnitude of the final velocity.

Momentum before collision is equal to the momentum after collision

m1v1+m2v2=(m1+m2)v

here, v is final velocity of the skateboarder, assuming right side velocity as positive, and left side velocity as negative.

70*12+2000*(-25)=(70+2000)v

v=-23.75 m/s

Here, negative sign indicates that both skyboarder and bus will move in left direction.

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Answer:

0.273 liters are needed to accomplish this task without boiling.

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If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

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$V_{k}$ - Volume of kerosene, measured in cubic meters.

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$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

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3 years ago