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Free_Kalibri [48]

2 years ago

4

A 2-oz bullet is fired horizontally with a velocity v1 = 1800 ft/sec into the 6-kg block of soft wood initially at rest on the h

orizontal surface. The bullet emerges from the block with the velocity v2 = 1200 ft/sec, and the block is observed to slide a distance of 8 ft before coming to rest. Determine the coefficient of kinetic friction μk between the block and the supporting surface.

1 answer:

Mashcka [7]2 years ago

4 0

Answer:0.0623

Explanation:

Given

Initial velocity of bullet=1800 ft/sec

Final velocity of bullet=1200 ft/sec

mass of bullet =2 oz=56.699 gm

Now conserving momentum

$m_{bullet}u_i=m_{bullet}u_f+m_{block}v$

$56.699\times 1800=56.699\times 1200+6000\times v$

v=5.669 ft/sec[/tex]

Now applying equation of motion to get Coefficient of friction

$v_f^2-u_i^2=2as$

where $a=\mu _k\times g$

here $u_i=5.669$

and $v_f=0$

$5.669^2=2\times \mu _k\times g\times 8$

here $g=32.2 ft/s^2$

$\mu _k=0.0623$

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5 0

2 years ago

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vitfil [10]

Answer:

0.002925 m

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Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature

<h2>for aluminium</h2>

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2 years ago

Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o

earnstyle [38]

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2 years ago

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11111nata11111 [884]

Answer:

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2 years ago

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alisha [4.7K]

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2 years ago