The acceleration of the bullet.

Question

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Answer 1

Question

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

(b)

To determine

The force on the bullet.

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

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Answer 2

Question

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

(b)

To determine

The force on the bullet.

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

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Answer 3

Question

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

(b)

To determine

The force on the bullet.

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

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Answer 4

Question

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

(b)

To determine

The force on the bullet.

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

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Answer 5

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

(b)

To determine

The force on the bullet.

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

Answer 6

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

Answer 7

Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

Answer 8

(a)

Expert Solution

Answer to Problem 82P

The bullet has an acceleration of $3.20×104m/s2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

$v=v0+abΔt$ (I)

Here, $v$ is the final velocity of the bullet, $v0$ is the initial velocity of the bullet, $ab$ is the acceleration of the bullet and $Δt$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $ab$ .

$ab=v−v0Δt$

Conclusion:

Substitute $320m/s$ for $v$ , $0m/s$ for $v0$ and $0.01s$ for $Δt$ .

$ab=320m/s−0m/s0.01s=3.20×104m/s2$

Therefore, the acceleration of the bullet is $3.20×104m/s2$ .

Answer 13

(b)

To determine

The force on the bullet.

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

Answer 14

(b)

To determine

The force on the bullet.

Answer 15

(b)

Expert Solution

Answer to Problem 82P

The force on the bullet is $96.0N$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

$Fb=mab$

Here, $Fb$ is the force on the bullet, $m$ is the mass of the bullet and $ab$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00×10−3kg$ for $m$ and $3.20×104m/s2$ for $ab$ .

$Fb=(3.00×10−3kg)(3.20×104m/s2)=96.0N$

Therefore, the force on the bullet is $96.0N$ .

Answer 20

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

Answer 21

(c)

To determine

The magnitude of force exerted on the gun.

Answer 22

(c)

Expert Solution

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0N$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

$Fb=−Fr$

Here, $Fb$ is the force on the bullet and $Fr$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0N$ for $Fb$ .

$Fr=−96.0N$

Therefore, the magnitude of force exerted on the gun is $96.0N$ . The negative sign in the answer indicates that the reaction is in the opposite direction.

Answer 27

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

Answer 28

(d)

To determine

The acceleration experienced by the gun

Answer 29

(d)

Expert Solution

Answer to Problem 82P

The gun experiences an acceleration of $−18.5m/s2$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

$Fr=Mar$

Here, $Fb$ is the force on the gun, $M$ is the mass of the gun and $ar$ is the acceleration of the gun.

Conclusion:

Substitute $−96.0N$ for $Fr$ and $5.20kg$ for $M$ to solve for $ar$ .

$−96.0N=(5.20kg)arar=−96.0N5.20kg=−18.5m/s2$

Therefore, the acceleration experienced by the gun is $−18.5m/s2$ .

Answer 34

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

Answer 35

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

Answer 36

(e)

Expert Solution

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

$Mm=5.20kg3.00×10−3kg=1.73×103$ (II)

Write the ratio of the acceleration and substitute.

$abar=3.20×104m/s2−18.5m/s2=−1.73×103$ (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

Answer 41

Ch. 3.2 - Prob. 3.1CC Ch. 3.3 - Prob. 3.2CC Ch. 3.4 - Prob. 3.3CC Ch. 3.4 - Prob. 3.4CC Ch. 3.5 - Prob. 3.5CC Ch. 3.6 - Prob. 3.6CC Ch. 3.7 - Acceleration of a Skydiver Figure 3.27 shows a...Ch. 3 - Prob. 1Q Ch. 3 - Prob. 2Q Ch. 3 - Prob. 3Q

The acceleration of the bullet.

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The acceleration of the bullet.

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The force on the bullet.

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The magnitude of force exerted on the gun.

Answer to Problem 82P

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The acceleration experienced by the gun

Answer to Problem 82P

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The ratio of mass $m$ and $M$ to that of the acceleration.

Answer to Problem 82P

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Chapter 3 Solutions

The acceleration of the bullet.

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The acceleration of the bullet.

Answer to Problem 82P

Explanation of Solution

The force on the bullet.

Answer to Problem 82P

Explanation of Solution

The magnitude of force exerted on the gun.

Answer to Problem 82P

Explanation of Solution

The acceleration experienced by the gun

Answer to Problem 82P

Explanation of Solution

The ratio of mass m<m:math><m:mi>m</m:mi></m:math> and M<m:math><m:mi>M</m:mi></m:math> to that of the acceleration.

Answer to Problem 82P

Explanation of Solution

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Chapter 3 Solutions

The ratio of mass $m$ and $M$ to that of the acceleration.