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4vir4ik [10]
3 years ago
6

A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu

me the net braking force acting on it has magnitude 8.01×10−16 N and is directed opposite to its initial velocity. Over what distance must the force be applied? Watch your negative signs in this problem.
Physics
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

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in the given case ,

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For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of
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The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

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<span>Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:</span>

2 = k (2)^2

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The equation becomes:

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Now solving for d when t = 4:

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<span>
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8 0
3 years ago
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