(a) The terminal velocity of the person with parachute.

Question

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Answer 1

Question

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

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Answer 2

Question

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

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Answer 3

Question

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

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Answer 4

Question

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

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Answer 5

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

Answer 6

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

Answer 7

Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Answer 8

Expert Solution

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17m s-1$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Proportionality constant value

- $12.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−(mg)=0$

To increase the velocity to $20.0 ms−1.$

$Fdrag on person +(−W person)=0For terminal velocityFdrag on person +(−mg)=0cv2terminal+(( −50.0 kg)⋅( 9.80 ms −2 ))=0cv2terminal=686 kgms−2v2terminal=686 kgms −2cvterminal= 686 kgms −2 c kgm −1 vterminal=1c⋅(26.192)ms−1vterminal=1 18 ⋅(26.192)ms−1=6.17m s-1$

Conclusion:

Thus, the terminal velocity of the person with parachute is $vterminal=6.17m s-1$

Answer 13

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

Answer 14

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Answer 15

Expert Solution

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1$ is $0.2744 kg ms−1$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Mass of the person

- $70.0 kg$

Terminal velocity

- $50.0 ms−1$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

$Fdrag on person +(−Wperson)=0$

$Fdrag=cv2$

Calculation:

$Fdrag on person +(−W person)=0For terminal velocitycv2terminal−mg=0$

$c⋅( ( 50 ms −1 )2)+(( −70.0 kg)⋅( 9.80 ms −2 ))=0c=686 kgms −22500 m2s −2c=0.2744 kg ms−1$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744 kg ms−1$

Answer 20

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

(a) The terminal velocity of the person with parachute.

Concept explainers

Videos

(a)

The terminal velocity of the person with parachute.

Answer to Problem 55QAP

Explanation of Solution

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$

Answer to Problem 55QAP

Explanation of Solution

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Chapter 5 Solutions

(a) The terminal velocity of the person with parachute.

Concept explainers

Videos

(a) The terminal velocity of the person with parachute.

Answer to Problem 55QAP

Explanation of Solution

(b) The value of the proportionality constant c when the terminal velocity is 50.0 ms−1.<m:math><m:mrow><m:mn>50.0</m:mn><m:msup><m:mrow><m:mtext> ms</m:mtext></m:mrow><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow></m:msup><m:mo>.</m:mo></m:mrow></m:math>

Answer to Problem 55QAP

Explanation of Solution

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Chapter 5 Solutions

(a)

The terminal velocity of the person with parachute.

(b)

The value of the proportionality constant c when the terminal velocity is $50.0 ms−1.$