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Vikentia [17]
3 years ago
5

A bullet of mass 6.00 gg is fired horizontally into a wooden block of mass 1.20 kgkg resting on a horizontal surface. The coeffi

cient of kinetic friction between block and surface is 0.190. The bullet remains embedded in the block, which is observed to slide a distance 0.270 mm along the surface before stopping.
(a) What was teh initial speed of the bullet?(b) What kind of collision took place between the bullet and the block? Explain your answer comparing the values of the kinetic energy before and after the collision.(c) Calculate the impulse of the block from the moment just after the impact to his final position. Analyze the result.
Physics
1 answer:
icang [17]3 years ago
3 0

Answer:

A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.230 m along the surface before stopping.

Explanation:

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A tank containing 200 L of hydrogen gas at 0.0 Celsius is kept at 10 kPa. The pressure is raised to 95C, and the volume is decre
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Answer:

The new pressure of the gas is 15.40 kPa.

Explanation:

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Mathematically this law indicates that the quotient between pressure and temperature is constant:

\frac{P}{T}=k

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Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. Mathematically, this law says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

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Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T}=k

Studying an initial state 1 and a final state 2, it is fulfilled:

\frac{P1*V1}{T1}=\frac{P2*V2}{T2}

In this case:

  • P1= 10 kPa
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Replacing:

\frac{10 kPa*200 L}{273 K}=\frac{P2*175 L}{368 K}

Solving:

P2=\frac{368 K}{175 L} *\frac{10 kPa*200 L}{273 K}

P2= 15.40 kPa

<u><em>The new pressure of the gas is 15.40 kPa.</em></u>

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