The average power delivered to the protons.

Question

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Answer 1

Question

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

(b)

To determine

The force acting on the protons.

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

(c)

To determine

The velocity of protons just before hitting the target.

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the acceleration of each proton.

a = Fmp …… (4)

Here, a is the acceleration and mp is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

vf2 = 2 a S …… (5)

Here, vf is the final velocity and S is the displacement.

Calculation:

Substitute 1.07 × 10−15 N for F and 1.67 × 10−27 Kg for mp in expression (4).

a= 1.07 × 10−15 N1.67 × 10−27 Kg = 6.41 × 1011 m s-2

Substitute 6.41 × 1011 m s-2 for a and 1.5 m for S in expression (5).

vf2= 2 (6.41 × 1011 m s-2) (1.5 m )= 1.923 × 1012 m2 s-2

Simplify the above expression.

vf = 1.923 × 1012 m s-1 = 1.38 × 106 m s-1

Conclusion:

Thus, the speed of the protons just before hitting the target is 1.38 × 106 m s-1 .

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Answer 2

Question

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

(b)

To determine

The force acting on the protons.

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

(c)

To determine

The velocity of protons just before hitting the target.

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the acceleration of each proton.

a = Fmp …… (4)

Here, a is the acceleration and mp is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

vf2 = 2 a S …… (5)

Here, vf is the final velocity and S is the displacement.

Calculation:

Substitute 1.07 × 10−15 N for F and 1.67 × 10−27 Kg for mp in expression (4).

a= 1.07 × 10−15 N1.67 × 10−27 Kg = 6.41 × 1011 m s-2

Substitute 6.41 × 1011 m s-2 for a and 1.5 m for S in expression (5).

vf2= 2 (6.41 × 1011 m s-2) (1.5 m )= 1.923 × 1012 m2 s-2

Simplify the above expression.

vf = 1.923 × 1012 m s-1 = 1.38 × 106 m s-1

Conclusion:

Thus, the speed of the protons just before hitting the target is 1.38 × 106 m s-1 .

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Answer 3

Question

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

(b)

To determine

The force acting on the protons.

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

(c)

To determine

The velocity of protons just before hitting the target.

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the acceleration of each proton.

a = Fmp …… (4)

Here, a is the acceleration and mp is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

vf2 = 2 a S …… (5)

Here, vf is the final velocity and S is the displacement.

Calculation:

Substitute 1.07 × 10−15 N for F and 1.67 × 10−27 Kg for mp in expression (4).

a= 1.07 × 10−15 N1.67 × 10−27 Kg = 6.41 × 1011 m s-2

Substitute 6.41 × 1011 m s-2 for a and 1.5 m for S in expression (5).

vf2= 2 (6.41 × 1011 m s-2) (1.5 m )= 1.923 × 1012 m2 s-2

Simplify the above expression.

vf = 1.923 × 1012 m s-1 = 1.38 × 106 m s-1

Conclusion:

Thus, the speed of the protons just before hitting the target is 1.38 × 106 m s-1 .

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Answer 4

Question

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

(b)

To determine

The force acting on the protons.

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

(c)

To determine

The velocity of protons just before hitting the target.

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the acceleration of each proton.

a = Fmp …… (4)

Here, a is the acceleration and mp is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

vf2 = 2 a S …… (5)

Here, vf is the final velocity and S is the displacement.

Calculation:

Substitute 1.07 × 10−15 N for F and 1.67 × 10−27 Kg for mp in expression (4).

a= 1.07 × 10−15 N1.67 × 10−27 Kg = 6.41 × 1011 m s-2

Substitute 6.41 × 1011 m s-2 for a and 1.5 m for S in expression (5).

vf2= 2 (6.41 × 1011 m s-2) (1.5 m )= 1.923 × 1012 m2 s-2

Simplify the above expression.

vf = 1.923 × 1012 m s-1 = 1.38 × 106 m s-1

Conclusion:

Thus, the speed of the protons just before hitting the target is 1.38 × 106 m s-1 .

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Answer 5

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

(b)

To determine

The force acting on the protons.

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

(c)

To determine

The velocity of protons just before hitting the target.

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the acceleration of each proton.

a = Fmp …… (4)

Here, a is the acceleration and mp is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

vf2 = 2 a S …… (5)

Here, vf is the final velocity and S is the displacement.

Calculation:

Substitute 1.07 × 10−15 N for F and 1.67 × 10−27 Kg for mp in expression (4).

a= 1.07 × 10−15 N1.67 × 10−27 Kg = 6.41 × 1011 m s-2

Substitute 6.41 × 1011 m s-2 for a and 1.5 m for S in expression (5).

vf2= 2 (6.41 × 1011 m s-2) (1.5 m )= 1.923 × 1012 m2 s-2

Simplify the above expression.

vf = 1.923 × 1012 m s-1 = 1.38 × 106 m s-1

Conclusion:

Thus, the speed of the protons just before hitting the target is 1.38 × 106 m s-1 .

Answer 6

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

Answer 7

Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

P = n(Energy of each proton)

Here, P is the required power and n is the number of protons reaching the target per second.

Calculation:

Substitute 1 × 109 for n and 10 keV for energy of each protonin the above expression.

P= (1 × 109 s-1) (10 keV )= (109 s-1)((10 keV) (1000 eV) 1 keV) ( 1.6 × 10−19 J 1 eV ) = 1.6 × 10−6 J s-1 = 1.6 × 10−6 W

Conclusion:

Thus, the average power that must be delivered to the stream of protons is 1.6 × 10−6 W .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The force acting on the protons.

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

Answer 13

(b)

To determine

The force acting on the protons.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

W= (Energy of each proton) t …… (1)

Here, W is the work done and t is the time required to eject one proton.

Write the expression for the net force on the stream protons.

Fnet = WS …… (2)

Here, W is the work done, Fnet is the net force and S is the displacement.

Write the expression for the force on each proton.

F = Fnet(number of protons) …… (3)

Calculation:

Substitute 1 s for t and 1.6 × 10−6 W for energy of eachproton in expression(1).

W = (1.6 × 10−6 W) (1s)= 1.6 × 10−6 J

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute 1.6 × 10−6 W for W and 1.5 0 m for S in expression (2).

Fnet= 1.6 × 10−6 J1.50 m = 1.07 × 10−6 N

Substitute 1 × 109 for the number of protons in expression (3).

F= 1.07 × 10−6 N1 ×109 = 1.07 × 10−15 N

Conclusion:

Thus, the force applied to each protons is 1.07 × 10−15 N .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

The velocity of protons just before hitting the target.

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .

Formula Used:

Write the expression for the acceleration of each proton.

a = Fmp …… (4)

Here, a is the acceleration and mp is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

vf2 = 2 a S …… (5)

Here, vf is the final velocity and S is the displacement.

Calculation:

Substitute 1.07 × 10−15 N for F and 1.67 × 10−27 Kg for mp in expression (4).

a= 1.07 × 10−15 N1.67 × 10−27 Kg = 6.41 × 1011 m s-2

Substitute 6.41 × 1011 m s-2 for a and 1.5 m for S in expression (5).

vf2= 2 (6.41 × 1011 m s-2) (1.5 m )= 1.923 × 1012 m2 s-2

Simplify the above expression.

vf = 1.923 × 1012 m s-1 = 1.38 × 106 m s-1

Conclusion:

Thus, the speed of the protons just before hitting the target is 1.38 × 106 m s-1 .

Answer 19

(c)

To determine

The velocity of protons just before hitting the target.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given:

Each of the proton has a kinetic energy 10 keV . Per second 1 × 109 protons are reaching the target at the end of the 1.50 m long accelerator. Mass of the proton is 1.67 × 10−27 Kg .