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

Question

Want to see more full solutions like this?

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Fresnel lens: You would like to design a 25 mm diameter blazed Fresnel zone plate with a first-order power of +1.5 diopters. What is the lithography requirement (resolution required) for making this lens that is designed for 550 nm? Express your answer in units of μm to one decimal point. Fresnel lens: What would the power of the first diffracted order of this lens be at wavelength of 400 nm? Express your answer in diopters to one decimal point. Eye: A person with myopic eyes has a far point of 15 cm. What power contact lenses does she need to correct her version to a standard far point at infinity? Give your answer in diopter to one decimal point.

Paraxial design of a field flattener. Imagine your optical system has Petzal curvature of the field with radius p. In Module 1 of Course 1, a homework problem asked you to derive the paraxial focus shift along the axis when a slab of glass was inserted in a converging cone of rays. Find or re-derive that result, then use it to calculate the paraxial radius of curvature of a field flattener of refractive index n that will correct the observed Petzval. Assume that the side of the flattener facing the image plane is plano. What is the required radius of the plano-convex field flattener? (p written as rho )

3.37(a) Five free electrons exist in a three-dimensional infinite potential well with all three widths equal to \( a = 12 \, \text{Å} \). Determine the Fermi energy level at \( T = 0 \, \text{K} \). (b) Repeat part (a) for 13 electrons. Book: Semiconductor Physics and Devices 4th ed, NeamanChapter-3Please expert answer only. don't give gpt-generated answers, & please clear the concept of quantum states for determining nx, ny, nz to determine E, as I don't have much idea about that topic.

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Concept explainers

Videos

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

Answer to Problem 66QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions