The gravitational force exerted by the bowling balls on a ping pong ball

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Answer 1

Question

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

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Answer 2

Question

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

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Answer 3

Question

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

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Answer 4

Question

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

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Answer 5

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

Answer 6

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

Answer 7

Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Answer 8

Expert Solution & Answer

Answer to Problem 66QAP

Total gravitational force =−8.023×10−12N

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given info:

Mass of a bowling ball = 8.0kg

Radius = 11.0cm

Distance from point P = 1.0m

Mass of ping pong ball = 2.70g

Formula used:

F=−GMmR2

F= Gravitational force

G= Gravitational constant

M= Mass of bowling ball

R= Radius

m= Mass of ping pong ball

Calculation:

Radius of each bowling bowl,

x1=(1.0m)2+(0.11*2m)2=1.02m

x2=(1.0m)2+(0.11*4m)2=1.09m

x3=(1.0m)2+(0.11*6m)2=1.20m

Gravitational force of each bowling bowl,

F1y=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.00m)2=−1.44*10−12N

F2=F5=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.02m)2=−1.38*10−12N

F2y=F5y=F2x1=−−1.38*10−12N1.02m=−1.35*10−12N

F3=F6=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.09m)2=−1.21*10−12N

F3y=F6y=F3x2=−−1.21*10−12N1.09m=−1.11*10−12N

F4=F7=−GMmx12=−(6.67*10−11Nm2/kg2)*8.0kg*(2.70*10−3kg)(1.20m)2=−1.00*10−12N

F4y=F7y=F3x3=−−1.00*10−12N1.20m=−8.33*10−13N

Total gravitational force,

Fnet=F1y+F2y+F3y+F4y+F5y+F6y+F7y=−1.44*10−12N+((−1.35*10−12N−1.11*10−12N−8.33*10−13N)*2)Fnet=−8.023*10−12N

Conclusion:

Total gravitational force =−8.023×10−12N

Answer 12

Ch. 10 - Prob. 1QAP Ch. 10 - Prob. 2QAP Ch. 10 - Prob. 3QAP Ch. 10 - Prob. 4QAP Ch. 10 - Prob. 5QAP Ch. 10 - Prob. 6QAP Ch. 10 - Prob. 7QAP Ch. 10 - Prob. 8QAP Ch. 10 - Prob. 9QAP Ch. 10 - Prob. 10QAP

The gravitational force exerted by the bowling balls on a ping pong ball

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The gravitational force exerted by the bowling balls on a ping pong ball

Answer to Problem 66QAP

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