COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution
Check Mark

Answer to Problem 90P

The velocity of the satellite at the point C is 3.07km/s.

Explanation of Solution

Write the expression for velocity of the satellite.

  v=2π(r+h)T        (I)

Here, v is the velocity of the satellite, r is the radius of the Earth, h is the altitude of the satellite above the Earth’s surface, and T is the orbital period of the Earth.

Conclusion:

Substitute 6371km for r, 35800km for h, and 86400s for T in equation (I) to find v.

  v=2π(6371km+35800km)86400s=2(3.14)(6371km+35800km)86400s=3.07km/s

Therefore, the velocity of the satellite at the point C is 3.07km/s.

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution
Check Mark

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is 2.76km/s.

Explanation of Solution

Write the expression for average velocity of the satellite.

  vavg=ΔrΔt        (II)

Here, vavg is the average velocity, Δr is the change in position, and Δt is the time period.

Substitute (r+h)2 for Δr and T/4 for Δt in equation (II).

  vavg=(r+h)2T/4=4(r+h)2T

Conclusion:

Substitute 6371km for r, 35800km for h, and 86400s for T in above relation to find vavg.

  vavg=4(6371km+35800km)2(86400s)=2.76km/s

Therefore, the angle does the wheel rotate during the third 1.0s time interval is 450°.

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution
Check Mark

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is 0.201m/s2.

Explanation of Solution

Write the expression for average acceleration of the satellite.

  aavg=ΔvΔt        (III)

Here, aavg is the average acceleration, Δv is the change in velocity, and Δt is the time period.

Since, the average acceleration is in the same direction as,

  Δv=vBvA=(Δvx)2+(Δvy)2=v2+v2=v2

Substitute v2 for Δv and T/4 for Δt in equation (III).

  aavg=v2T/4=4v2T

Conclusion:

Substitute 3.07km/s for v and 86400s for T in above relation to find aavg.

  aavg=4(3.07km/s×103m1km)2(86400s)=4(3.07×103m/s)2(86400s)=0.201m/s2

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is 0.201m/s2.

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution
Check Mark

Answer to Problem 90P

The acceleration of the satellite at point D is 0.224m/s2.

Explanation of Solution

Write the expression for gravitational force.

  F=GmMEr2        (IV)

Here, F is the gravitational force exerted on the satellite by the Earth, G is the gravitational constant, m is the mass of the satellite, ME is the mass of the Earth, and r is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

  F=mac        (V)

Here, F is the force exerted on the satellite and ac is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

  mac=GmMEr2ac=GME(r+h)2

Substitute 6.673×1011Nm2/kg2 for G, 5.97×1024kg for ME, 6371km for r, and 35800km for h in above relation to find ac.

  ac=(6.673×1011Nm2/kg2)(5.97×1024kg)(6371km×103m1km+35800km×103m1km)2=(6.673×1011Nm2/kg2)(5.97×1024kg)(6371×103m+35800×103m)2=0.224m/s2

Therefore, the acceleration of the satellite at point D is 0.224m/s2.

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Chapter 5 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 5.4 - Prob. 5.8PPCh. 5.4 - Prob. 5.4CPCh. 5.4 - Prob. 5.9PPCh. 5.4 - Prob. 5.10PPCh. 5.5 - Prob. 5.5CPCh. 5.5 - Prob. 5.11PPCh. 5.5 - Prob. 5.12PPCh. 5.6 - Prob. 5.6CPCh. 5.6 - Prob. 5.13PPCh. 5.7 - Prob. 5.14PPCh. 5 - Prob. 1CQCh. 5 - Two children ride on a merry-go-round. One is 2 m...Ch. 5 - Explain why the orbital radius and the speed of a...Ch. 5 - In uniform circular motion, is the velocity...Ch. 5 - In uniform circular motion, the net force is...Ch. 5 - The speed of a satellite in circular orbit around...Ch. 5 - A flywheel (a massive disk) rotates with constant...Ch. 5 - Explain why the force of gravity due to Earth does...Ch. 5 - When a roller coaster takes a sharp turn to the...Ch. 5 - Is there anywhere on Earth where a bathroom scale...Ch. 5 - A physics teacher draws a cutaway view of a car...Ch. 5 - A bridal party is at a rehearsal dinner. The best...Ch. 5 - A leopard starts from rest at t = 0 and runs in a...Ch. 5 - What is the direction of the satellite’s...Ch. 5 - Questions 1–4: A satellite in orbit travels around...Ch. 5 - What is the direction of the satellite’s...Ch. 5 - An object moving in a circle at a constant speed...Ch. 5 - A spider sits on a DVD that is rotating at a...Ch. 5 - Two satellites are in orbit around Mars with the...Ch. 5 - Questions 8–9: A boy swings in a tire swing....Ch. 5 - Prob. 9MCQCh. 5 - Prob. 10MCQCh. 5 - Prob. 11MCQCh. 5 - Prob. 12MCQCh. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Dung beetles are renowned for building large...Ch. 5 - Prob. 9PCh. 5 - Prob. 10PCh. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - A child’s toy has a 0.100 kg ball attached to two...Ch. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - A curve in a stretch of highway has radius 512 m....Ch. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - A car drives around a curve with radius 410 m at a...Ch. 5 - Prob. 28PCh. 5 - A road with a radius of 75.0 m is banked so that...Ch. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - A person of mass M stands on a bathroom scale...Ch. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 64PCh. 5 - Prob. 65PCh. 5 - Prob. 66PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - A jogger runs counterclockwise around a path of...Ch. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 81PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - In Chapter 19 we will see that a charged particle...Ch. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94P
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