MODERN PHYSICS F/SCI..-WEBASSIGN(MULTI)
MODERN PHYSICS F/SCI..-WEBASSIGN(MULTI)
4th Edition
ISBN: 9780357644782
Author: Thornton
Publisher: CENGAGE L
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Chapter 2, Problem 15P

<a>

To determine

Relativistic factor for speed 95 km / h.

<a>

Expert Solution
Check Mark

Answer to Problem 15P

The relativistic factor for speed 95 km / h. is 1+3.86×1015.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

γ=11v2c2=11(vc)2=11β2 

Because β=vc, where, v is the speed of the object and c is the speed of the light.

Convert, speed kilometer per hour to meter per second. Thus

v=95 km / h=95 km / h×(1000m1.0km)(1.0h3600s)=95036 m / s=26.38 m / s

Here, v<<c, thus β=vc will be very small, therefore, for smaller value of β use binomial expansion for equation γ=11β2. Therefore,

γ=11β2γ=1(1β2)12γ=(1β2)12γ=1(12)×β2

Further simplify the above equation. Thus,

γ=1(12)×β2γ=1+β22

Substitute the value of the β=vc in the above equation. Thus,

γ=1+(vc)22γ=1+12(vc)2

Substitute v=26.38 m / s and c=3×108 m / s in above equation to calculate relativistic factor. Thus,

γ=1+12(26.38 m / s3×108 m / s)2=1+12(8.79×108)2=1+12×77.26×1016=1+3.86×1015

Conclusion:

Therefore, relativistic factor for speed 95 km / h. is 1+3.86×1015.

<b>

To determine

Relativistic factor for v=240 m / s.

<b>

Expert Solution
Check Mark

Answer to Problem 15P

The relativistic factor for v=240 m / s is 1+3.2×1013.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

γ=11v2c2=11(vc)2=11β2 

Because β=vc, where, v is the speed of the object and c is the speed of the light.

Here, v=240 m / s which leads v<<c, thus β=vc will be very small, therefore, for smaller value of β use binomial expansion for equation γ=11β2. Therefore,

γ=11β2γ=1(1β2)12γ=(1β2)12γ=1(12)×β2

Further simplify the above equation. Thus,

γ=1(12)×β2γ=1+β22

Substitute the value of the β=vc in the above equation. Thus,

γ=1+(vc)22γ=1+12(vc)2

Substitute v=240 m / s and c=3×108 m / s in above equation to calculate relativistic factor. Thus,

γ=1+12(240 m / s3×108 m / s)2=1+12(8.0×107)2=1+12×64×1014=1+3.2×1013

Conclusion:

Therefore, relativistic factor for v=2.3 Mach is 1+3.2×1013.

<c>

To determine

The ratio β=vc, for an airplane travelling with 2.3 Mach.

<c>

Expert Solution
Check Mark

Answer to Problem 15P

The relativistic factor for an airplane travelling with 2.3 Mach is 3.2×1012.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

γ=11v2c2=11(vc)2=11β2 

Because β=vc, where, v is the speed of the object and c is the speed of the light.

Velocity of the sound in the is vs=330 m / s.

Write the formula for the conversion of the speed of the supersonic plane from Mach to meter per second. Thus,

v=Mvs, where, vs is the velocity of the sound in the air and M is the Mach number.

Substitute vs=330 m / s and M=2.3 in in v=Mvs to  get v. Thus,

v=2.3×330 m / s=759 m / s

Here, v=759 m / s which leads v<<c, thus β=vc will be very small, therefore, for smaller value of β use binomial expansion for equation γ=11β2. Therefore,

γ=11β2γ=1(1β2)12γ=(1β2)12γ=1(12)×β2

Further simplify the above equation. Thus,

γ=1(12)×β2γ=1+β22

Substitute the value of the β=vc in the above equation. Thus,

γ=1+(vc)22γ=1+12(vc)2

Substitute v=759 m / s and c=3×108 m / s in above equation to calculate relativistic factor. Thus,

γ=1+12(759 m / s3×108 m / s)2=1+12(253×108)2=1+12(2.53×106)2=1+12×6.4×1012,

Simplify the above equation. Thus,

γ=1+12×6.4×1012=1+3.2×1012

Conclusion:

Therefore, the relativistic factor for speed 2.3 Mach is 1+3.2×1012.

<d>

To determine

The relativistic factor for v=27000 km / h.

<d>

Expert Solution
Check Mark

Answer to Problem 15P

The relativistic factor for v=27000 km / h is 1+3.13×1010.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

γ=11v2c2=11(vc)2=11β2 

Because β=vc, where, v is the speed of the object and c is the speed of the light.

Convert, speed kilometer per hour to meter per second. Thus

v=27000 km / h=27000 km / h×(1000m1.0km)(1.0h3600s)=270×10336 m / s=7.5×103 m / s

Here, v=7.5×103 m / s which leads v<<c, thus β=vc will be very small, therefore, for smaller value of β use binomial expansion for equation γ=11β2. Therefore,

γ=11β2γ=1(1β2)12γ=(1β2)12γ=1(12)×β2

Further simplify the above equation. Thus,

γ=1(12)×β2γ=1+β22

Substitute the value of the β=vc in the above equation. Thus,

γ=1+(vc)22γ=1+12(vc)2

Substitute v=7.5×103 m / s and c=3×108 m / s in above equation to calculate relativistic factor. Thus,

γ=1+12(7.5×103 m / s3×108 m / s)2=1+12(2.5×105)2=1+12×6.25×1010=1+3.13×1010,

Conclusion:

Therefore, the relativistic factor for v=27000 km / h is 1+3.13×1010.

<e>

To determine

The relativistic factor, when an electron covering a distance 25 cm in 2 ns.

<e>

Expert Solution
Check Mark

Answer to Problem 15P

The relativistic factor, when an electron covering a distance 25 cm in 2 ns is 1.10.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

γ=11v2c2=11(vc)2=11β2 

Because β=vc, where, v is the speed of the object and c is the speed of the light.

Write the equation to calculate the speed of the electron. Thus

v=dt, where, d is the distance covered in time t.

Distance covered by the electron is 25 cm. Convert this distance in meter, thus,

25 cm=25.0 cm×102 m1.0 cm=25.0×102 m

Time taken by the electron to cover 25 cm is 2 ns. Convert this time in second, thus,

2 ns=2 ns×109 s1 ns=2×109 s

Substitute d=25.0×102 m and t=2×109 s in v=dt to calculate the v. Thus,

v=25.0×102 m2×109 s=12.5×107 m / s=1.25×108 m / s

Here, v=1.25×108 m / s is comparable to speed of light c. Thus β=vc will not be very small, therefore, calculate the β=vc, by substituting the value of the v and c.

Substitute v=1.25×108 m / s in β=vc, where, c=3×108 m / s is the speed of the light. Thus,

β=1.25×108 m / s3×108 m / s=0.416 =0.42 

Substitute β=0.42 in equation γ=11β2 to calculate the relativistic factor. Thus,

γ=11(0.42)2=1.10

Conclusion:

Therefore, the relativistic factor for v=1.25×108 m / s is 1.10.

<f>

To determine

The relativistic factor, when an electron covering a distance 1014 m in 3.5×1023 s.

<f>

Expert Solution
Check Mark

Answer to Problem 15P

The relativistic factor, when an electron covering a distance 1014 m in 3.5×1023 s is 3.2.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

γ=11v2c2=11(vc)2=11β2 

Because β=vc, where, v is the speed of the object and c is the speed of the light.

Write the equation to calculate the Speed of the electron. Thus

v=dt, where, d is the distance covered in time t.

Distance covered by the electron is 1014 m and time taken to cover this distance is 3.5×1023 s.

Substitute d=1014 m and t=3.5×1023 s in v=dt to calculate the v. Thus,

v=1014 m3.5×1023 s=0.285×1014+23 m / s=0.285×109 m / s=2.86×108 m / s

Here, v=2.86×108 m / s is comparable to speed of light c. Thus β=vc will not be very small, therefore, calculate the β=vc, by substituting the value of the v and c.

Substitute v=2.86×108 m / s in β=vc, where, c=3×108 m / s is the speed of the light. Thus,

β=2.86×108 m / s3×108 m / s=0.95 

Substitute β=0.95 in equation γ=11β2 to calculate the relativistic factor. Thus,

γ=11(0.95)2=3.2

Conclusion:

Therefore, the relativistic factor for v=2.86×108 m / s is 3.2.

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Chapter 2 Solutions

MODERN PHYSICS F/SCI..-WEBASSIGN(MULTI)

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