Equation for conservation of energy.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

(c)

To determine

The total mass of final particle.

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

(d)

To determine

The difference between the masses.

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

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

Problem 04.08 (17 points). Answer the following questions related to the figure below. ථි R₁ www R₂ E R₁ www ли R₁ A Use Kirchhoff's laws to calculate the currents through each battery and resistor in terms of R1, R2, E1, & E2. B Given that all the resistances and EMFs have positive values, if E₁ > E2 and R₁ > R2, which direction is the current flowing through E₁? Through R₂? C If E1 E2 and R₁ > R2, which direction is the current flowing through E₁? Through R2?

A 105- and a 45.0-Q resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.268 A. When the 45.0-resistor is disconnected, the current from the battery drops to 0.0840 A. Determine (a) the emf and (b) the internal resistance of the battery. 10 R2 R₁ ww R₁ Emf 14 Emf Final circuit Initial circuit

A ball is shot at an angle of 60° with the ground. What should be the initial velocity of the ball so that it will go inside the ring 8 meters away and 3 meters high. Suppose that you want the ball to be scored exactly at the buzzer, determine the required time to throw and shoot the ball. Full solution and figure if there is.

Answer 2

Question

Chapter 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

(c)

To determine

The total mass of final particle.

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

(d)

To determine

The difference between the masses.

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

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 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

(c)

To determine

The total mass of final particle.

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

(d)

To determine

The difference between the masses.

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

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 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

(c)

To determine

The total mass of final particle.

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

(d)

To determine

The difference between the masses.

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

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 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

(c)

To determine

The total mass of final particle.

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

(d)

To determine

The difference between the masses.

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

Answer 6

Chapter 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Answer 7

Chapter 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

Answer 8

(a)

Expert Solution

Answer to Problem 36P

The equation for conservation of energy is

$K.Ei+mic2=K.Ef+mfc2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The formula used to calculate the initial rest energy is,

$ERi=mic2$

$ERi$ is the initial rest energy
$c$ is the speed of light
$mi$ is the initial mass

The formula used to calculate the final relativistic energy is,

$Ei=K.Ei+ERi=K.Ei+mic2$

The formula used to calculate the final rest energy is,

$ERf=mfc2$

$ERf$ is the final rest energy
$mf$ is the final mass

The formula used to calculate the final relativistic energy is,

$Ef=K.Ef+ERf=K.Ef+mfc2$

The formula used to calculate the conservation energy is,

$Ei=EfK.Ei+mic2=K.Ef+mfc2$

Thus, the equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Conclusion:

The equation for conservation of energy is $K.Ei+mic2=K.Ef+mfc2$ .

Answer 13

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

Answer 14

(b)

To determine

The total mass of initial particle.

Answer 15

(b)

Expert Solution

Answer to Problem 36P

The total mass of initial particle is

$236.052588 u$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info:

Mass of $U92235$ is $235.043923 u$ .

Mass of $n01$ is $1.008665 u$ .

The formula used to calculate the mass of initial particle is,

$mi=mU92235+mn01$

$mi$ is the mass of initial particle
$mU92235$ is the mass of $u92235$
$mn01$ is the mass of $n01$

Substitute $235.043923 u$ for $mU92235$ and $1.008665 u$ for $mn01$ to find $mi$ .

$mi=235.043923 u+1.008665 u=236.052588 u$

Thus, the total mass of initial particle is $236.052588 u$ .

Conclusion:

The total mass of initial particle is $236.052588 u$ .

Answer 20

(c)

To determine

The total mass of final particle.

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

Answer 21

(c)

To determine

The total mass of final particle.

Answer 22

(c)

Expert Solution

Answer to Problem 36P

The total mass of final particle is

$235.861612 u$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info:

Mass of $L57148a$ is $147.932236 u$ .

Mass of $B3587r$ is $86.9207119 u$ .

The formula used to calculate the mass of initial particle is,

$mf=mL57148a+mB3587r+mn01$

$mf$ is the mass of initial particle
$mL57148a$ is the mass of $L57148a$
$mB3587r$ is the mass of $B3587r$

Substitute $147.932236 u$ for $mL57148a$ , $86.9207119 u$ for $mB3587r$ and $1.008665 u$ for $mn01$ to find $mf$ .

$mf=147.932236 u+86.9207119 u+1.008665 u=235.861612 u$

Thus, the total mass of final particle is $235.861612 u$ .

Conclusion:

The total mass of final particle is $235.861612 u$ .

Answer 27

(d)

To determine

The difference between the masses.

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

Answer 28

(d)

To determine

The difference between the masses.

Answer 29

(d)

Expert Solution

Answer to Problem 36P

The mass difference is

$0.190976 u$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The formula used to calculate the mass difference is,

$Δm=mi−mf$

$Δm$ is the mass is the mass difference

Substitute $236.052588 u$ for $mi$ and $235.861612 u$ for $mf$ to find $Δm$ .

$Δm=236.052588 u−235.861612 u=0.190976 u$

Thus, the mass difference is $0.190976 u$ .

Conclusion:

The mass difference is $0.190976 u$ .

Answer 34

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

Answer 35

(e)

To determine

The final kinetic energy in MeV.

Answer 36

(e)

Expert Solution

Answer to Problem 36P

The final kinetic energy in MeV is

$177.893 MeV$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$K.Ef=Δm=(0.190976 u)(931.494 MeV1 u)=177.893 MeV$

Thus, the final kinetic energy in MeV is $177.893 MeV$ .

Conclusion:

The final kinetic energy in MeV is $177.893 MeV$

Answer 41

Ch. 26.3 - Prob. 26.1QQ Ch. 26.4 - Suppose youre an astronaut being paid according to...Ch. 26.4 - True or False: People traveling near the speed of...Ch. 26.4 - You are packing for a trip to another star, and on...Ch. 26.4 - You observe a locket moving away from you. (i)...Ch. 26.7 - Prob. 26.6QQ Ch. 26.7 - Prob. 26.7QQ Ch. 26 - Choose the option from each pair that makes the...Ch. 26 - Choose the option that makes the following...Ch. 26 - Choose the option that makes the following...

Equation for conservation of energy.

Videos

Equation for conservation of energy.

Answer to Problem 36P

Explanation of Solution

The total mass of initial particle.

Answer to Problem 36P

Explanation of Solution

The total mass of final particle.

Answer to Problem 36P

Explanation of Solution

The difference between the masses.

Answer to Problem 36P

Explanation of Solution

The final kinetic energy in MeV.

Answer to Problem 36P

Explanation of Solution

Want to see more full solutions like this?

Chapter 26 Solutions