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College Physics (Urone)
Found in: Page 82
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of \({\bf{6}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) for\({\bf{8}}{\bf{.10 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{s}}\) . What is its muzzle velocity (that is, its final velocity)?

The muzzle velocity of bullet is \(502.2\;{\rm{m/s}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Determination of formula for muzzle velocity of bullet

Given Data:

The initial velocity of bullet is \(u = 0\)

The time for acceleration of bullet is \(t = 8.10 \times {10^{ - 4}}\;{\rm{s}}\)

The acceleration for bullet is \(a = 6.20 \times {10^5}\;{\rm{m}}/{{\rm{s}}^2}\)

The muzzle velocity of the bullet is found by using the first equation of motion and by considering the initial velocity of bullet as zero.

The muzzle velocity of ball is given as

\(v = u + at\)

Here, \(v\) is the muzzle velocity of bullet.

Determination of muzzle velocity of bullet

Substitute all the values in the above equation.

\(\begin{array}{l}v = 0 + \left( {6.20 \times {{10}^5}\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {8.10 \times {{10}^{ - 4}}\;{\rm{s}}} \right)\\v = 502.2\;{\rm{m}}/{\rm{s}}\end{array}\)

Therefore, the muzzle velocity of bullet is \(502.2\;{\rm{m}}/{\rm{s}}\).

Most popular questions for Physics Textbooks

Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later.

(a) How far above the hiker is the rock when he can see it?

(b) How much time does he have to move before the rock hits his head?

a) Calculate the height of a cliff if it takes \({\bf{2}}.{\bf{35}}{\rm{ }}{\bf{s}}.\) for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of\({\bf{8}}.{\bf{00}}{\rm{ }}{\bf{m}}/{\bf{s}}\). (b) How long would it take to reach the ground if it is thrown straight down with the same speed?

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (\({\bf{8}}{\bf{.00 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}\;{\bf{s}}\)) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

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