College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 26, Problem 43AP

An astronaut wishes to visit the Andromeda galaxy, making a one-way trip that will take 30.0 years in the space-ship’s frame of reference. Assume the galaxy is 2.00 million light-years away and his speed is constant. (a) How fast must he travel relative to Earth? (b) What will be the kinetic energy of his spacecraft, which has mass of 1.00 × 106 kg? (c) What is the cost of this energy if it is purchased at a typical consumer price for electric energy, 13.0 cents per kWh? The following approximation will prove useful:

1 1 + x 1 x 2  for  x 1

(a)

Expert Solution
Check Mark
To determine
The speed of the observer relative to Earth.

Answer to Problem 43AP

The speed of the observer relative to Earth is 11.13×1010 .

Explanation of Solution

Given Info:

The distance is 2.00×106ly .

The time taken to travel is 30.0yr .

Formula to calculate the time is,

Δt=Δtp1(vc)2

  • Δtp is the time measured by the observer in spacecraft to cover the distance
  • Δt is time measured by the observer at Earth
  • c is speed of light
  • v is the speed of spacecraft

Formula to calculate the speed is,

v=dΔt=dΔtp1(vc)2

  • d is the distance covered

Substitute 2.00×106ly for d and 30.0yr for Δt to find v .

v=dΔt=(2.00×106ly)(1cly)30.0yr1(vc)2(1.50×105)(vc)=1(vc)2vc12.25×10102=11.13×1010

Thus, the speed of the observer relative to Earth is 11.13×1010 .

Conclusion:

The speed of the observer relative to Earth is 11.13×1010 .

(b)

Expert Solution
Check Mark
To determine
The kinetic energy of the spacecraft.

Answer to Problem 43AP

The kinetic energy of the spacecraft is 5.99×1027J .

Explanation of Solution

Given Info:

The mass of spacecraft is 1.00×106kg .

The speed of light is 3.00×108m/s .

Formula to kinetic energy is,

K.E=(γ1)mc2=(11(vc)21)mc2

  • m is the mass of spacecraft
  • K.E is the kinetic energy

Substitute 11.13×1010 for v/c , 1.00×106kg for m and 3.00×108m/s for c to find K.E .

K.E=(11(11.13×1010)21)(1.00×106kg)(3.00×108m/s)2=5.99×1027J

Thus, the kinetic energy of the spacecraft is 5.99×1027J .

Conclusion:

The kinetic energy of the spacecraft is 5.99×1027J .

(c)

Expert Solution
Check Mark
To determine
The cost of energy.

Answer to Problem 43AP

The cost of energy is $2.16×1020 .

Explanation of Solution

Given Info:

The rate per kilowatt is $0.13/kWh .

Formula to calculate cost of energy is,

C=R(K.E)

  • C is the cost of energy
  • R is the rate per kilowatt

Substitute 5.99×1027J for K.E and $0.13/kWh for R to find C .

C=($0.13/kWh)(5.99×1027J)(1kWh3.60×106J)=$2.16×1020

Thus, the cost of energy is $2.16×1020 .

Conclusion:

The cost of energy is $2.16×1020

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Chapter 26 Solutions

College Physics:

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