(a) Einsteinian momentum of an electron travels at 0.444c?

Question

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Answer 1

Question

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

To determine

(c)

Rest energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

To determine

(d)

The total energy of the electrontravels at 0.444c?

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

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Answer 2

Question

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

To determine

(c)

Rest energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

To determine

(d)

The total energy of the electrontravels at 0.444c?

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

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Answer 3

Question

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

To determine

(c)

Rest energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

To determine

(d)

The total energy of the electrontravels at 0.444c?

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

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

Answer 4

Question

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

To determine

(c)

Rest energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

To determine

(d)

The total energy of the electrontravels at 0.444c?

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

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Answer 5

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

To determine

(c)

Rest energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

To determine

(d)

The total energy of the electrontravels at 0.444c?

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

Answer 6

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

Answer 7

Chapter 25, Problem 65QAP

To determine

(a)

Einsteinian momentum of an electron travels at 0.444c?

Answer 8

Expert Solution

Answer to Problem 65QAP

Einsteinian momentum =1.35×10−22kgm/s

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

An electron (mass, m=9.11×10−31kg ) travels at v = 0.444c. The magnitude of theEinsteinian momentum is p=γmv, where γ=11− v2 c2 , The total energy E of the electronis the sum of its Einsteinian kinetic energy Kand the energy not associated with its motion,i.e., its rest energy, E0. The rest energy is equal to E0=mc2, and its total energy is equal to E=γmc2.

Relativistic gamma:

γ=11− v2 c2 =11− (0.444c)2 c2 =1.116

p=γmv=1.116×9.11×10−31×0.444×3×108=1.35×10−22kgm/s

Conclusion:

Einsteinian momentum =1.35×10−22kgm/s

Answer 13

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

Answer 14

To determine

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Answer 15

Expert Solution

Answer to Problem 65QAP

Einsteinian kinetic energy =9.51×10−15J

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

K+E0=EK=E−E0=γmc2−mc2=(γ−1)mc2=(1.116−1)×9.11×10−31×(3×108)2=9.51×10−15J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Einsteinian kinetic energy =9.51×10−15J

Answer 20

To determine

(c)

Rest energy of an electron travels at 0.444c?

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

Answer 21

To determine

(c)

Rest energy of an electron travels at 0.444c?

Answer 22

Expert Solution

Answer to Problem 65QAP

Rest energy =8.20×10−14J

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

E0=mc2=9.11×10−31×(3×108)2=8.20×10−14J

Einsteinian kinetic energy =9.51×10−15J

Conclusion:

Rest energy =8.20×10−14J

Answer 27

To determine

(d)

The total energy of the electrontravels at 0.444c?

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

Answer 28

To determine

(d)

The total energy of the electrontravels at 0.444c?

Answer 29

Expert Solution

Answer to Problem 65QAP

The total energy =9.15×10−14J

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

E=γmc2=1.116×9.11×10−31×(3×108)2=9.15×10−14J

The total energy =9.15×10−14J

Conclusion:

The total energy =9.15×10−14J

Answer 34

Ch. 25 - Prob. 1QAP Ch. 25 - Prob. 2QAP Ch. 25 - Prob. 3QAP Ch. 25 - Prob. 4QAP Ch. 25 - Prob. 5QAP Ch. 25 - Prob. 6QAP Ch. 25 - Prob. 7QAP Ch. 25 - Prob. 8QAP Ch. 25 - Prob. 9QAP Ch. 25 - Prob. 10QAP

(a) Einsteinian momentum of an electron travels at 0.444c?

Videos

(a)

Einsteinian momentum of an electron travels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

(c)

Rest energy of an electron travels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

(d)

The total energy of the electrontravels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 25 Solutions

(a) Einsteinian momentum of an electron travels at 0.444c?

Videos

(a) Einsteinian momentum of an electron travels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

(b) Einsteinian kinetic energy of an electron travels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

(c) Rest energy of an electron travels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

(d) The total energy of the electrontravels at 0.444c?

Answer to Problem 65QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 25 Solutions

(a)

Einsteinian momentum of an electron travels at 0.444c?

(b)

Einsteinian kinetic energy of an electron travels at 0.444c?

(c)

Rest energy of an electron travels at 0.444c?

(d)

The total energy of the electrontravels at 0.444c?