COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 6, Problem 103QAP
To determine

(a)

The speed of the material as it leaves the surface.

Expert Solution
Check Mark

Answer to Problem 103QAP

The speed of the material as it leaves the surface is 23.56m/sec.

Explanation of Solution

Given:

  gmars=3.7m/sec2h=75.0m

Concept Used:

Law of conservation of mechanical energy.

Calculation:

From the law of conservation of mechanical energy, we can calculate the speed of the material as it leaves the surface. The gravitational potential energy of the material at the surface is to be zero.

As the gas is travelling just below the surface of the surface of the planet, we can ignore the change in gravitational potential energy

According to law of conservation of energy:

  Ui+Ki=Uf+Kf0+12mvi2=mgmarsh+0vi=2gmarsh

On replacing the values, we get

  gmars=3.7m/sec2h=75.0mvi=2×3.7×75.0vi=23.56m/sec

Conclusion:

The speed of the material as it leaves the surface is 23.56m/sec.

To determine

(b)

The energy per kilogram of material is lost due to nonconservative forces.

Expert Solution
Check Mark

Answer to Problem 103QAP

The energy per kilogram of material is lost due to nonconservative forces is 7.0×102J/kg.

Explanation of Solution

Given:

  gmars=3.7m/sec2h=75.0mvi=160km/hr

Concept Used:

Law of nonconservative of work.

Calculation:

From the law of conservation of mechanical energy, we can calculate the speed of the material as it leaves the surface. The gravitational potential energy of the material at the surface is to be zero.

As the gas is travelling just below the surface of the surface of the planet, we can ignore the change in gravitational potential energy.

As the gas leaves the jets it is just travelling underground at the speed of 160km/hr

So, vi=160km/hr

Converting the units:

  vi=160km/hr×103m1km×1h3600secvi=44.4m/sec

According to law of conservation of energy:

  Ui+Ki=Uf+Kfmgmarsh+0=0+12mvf2vf=2gmarsh

On replacing the values, we get

  gmars=3.7m/sec2h=75.0mvf=2×3.7×75.0vf=23.56m/sec

According to law of nonconservative of work:

  Wnonconservative=ΔK+ΔUWnonconservative=12m(vf2vi2)+0E=WmE=Wnonconservativem=12(vf2vi2)

On replacing the values, we get

  vi=44.4m/secvf=23.56m/secE=12(23.56244.42)E=7.0×102J/kg

Conclusion:

The energy per kilogram of material is lost due to nonconservative forces is 7.0×102J/kg.

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Chapter 6 Solutions

COLLEGE PHYSICS

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