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3 years ago

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A 0.0121-kg bullet is fired straight up at a falling wooden block that has a mass of 4.99 kg. The bullet has a speed of 898 m/s

when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

1 answer:

aliya0001 [1]3 years ago

7 0

Answer:

t = 0.11 s

Explanation:

The motion is complete inelastic so the motion can be model using

mb * vb + mB * vB = ( mb + mB ) vf

mb = 0.0121 kg , mB = 4.99 kg , vb = 898 m / s , vB = 0

Replacing to find vf

0.0121 * 898 m / s + 4.99 kg * 0 = ( 0.0121 + 4.99) * vf

vf = 2.18 m / s

Now to find the time take the acceleration as a= g = 9.8 m /s²

t = mb * vb / a * (mb + 2 * mB)

t = [0.0121 kg * 898 m /s] / 9.8 * ( 0.0121 kg + 2*4.99)

t = 0.11 s

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A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i

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$T_f=5854.76$ °C

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<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

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Energy produced $= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

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$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$ .................(1)

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$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

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