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gayaneshka [121]
2 years ago
14

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average forc

e (in N) exerted on a 0.0200 kg bullet to accelerate it to a speed of 700 m/s in a time of 1.50 ms (milliseconds)
Physics
1 answer:
Natali5045456 [20]2 years ago
7 0

Answer:

<h2>9.3kN</h2>

Explanation:

Step one:

given data

mass of bullet= 0.02kg

speed v=700m/s

time taken =1.5ms= 0.0015 seconds

Step two:

we know that from the first law

F=ma-----1  first law of motion

also, we know that

a=v/t----2

put a=v/t in equation 1 we have

F=mv/t

Step three:

substitute our given data to find force

F=0.02*700/0.0015

F=14/0.0015

F=9333.33N

F=9.3kN

<u>The average force exerted is 9.3kN</u>

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A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin
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Answer:

U_k = 0.113

Explanation:

using the law of the conservation of energy:

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\frac{1}{2}Kx^2=NU_kd

where K is the spring constant, x is the spring compression, N is the normal force of the block, U_k is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

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So, Replacing values on the first equation, we get:

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Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
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Answer:

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Explanation:

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a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

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a_r=r''-r\theta'^2

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The radial component of acceleration is 8.77 m/s^2

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