COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 5, Problem 92P

(a)

To determine

The angular displacement in radians in an elapsed time of 1 day.

(a)

Expert Solution
Check Mark

Answer to Problem 92P

The angular displacement in radians in an elapsed time of 1 day is 0.0172rad_.

Explanation of Solution

Write the expression for the angular displacement.

    Δθ=ωΔt        (I)

Here, Δθ is the angular displacement, ω is the angular velocity, t is the time.

Write the expression for the angular speed.

    ω=2πT        (II)

Conclusion:

Substitute 365.25days for

   ω=2π365.25days×(24h1day)(3600s1h)=2π31557600s=1.99×107s

Substitute 1day for Δt and S for ω in equation (I) to find Δθ.

    Δθ=1.99×107s(1day×24h1day)(3600s1h)=0.0172rad

Therefore, angular displacement in radians in an elapsed time of 1 day is 0.0172rad_.

(b)

To determine

The change in Earths velocity.

(b)

Expert Solution
Check Mark

Answer to Problem 92P

The change in Earths velocity is 513.5m/s_ toward the center of Sun.

Explanation of Solution

Write the expression for the velocity.

    v=rω        (III)

Here, ω is the angular velocity, r is the radius.

The magnitude of change in velocity in 1 day is,

    |Δv|=vΔθ        (IV)

Conclusion:

Substitute 1.5×1011m for r, 1.99×107rad/s for ω in equation (III) to find v.

    v=1.5×1011m×(1.99×107rad/s)=2.98×104m/s

Substitute 2.98×104m/s for v and 0.0172rad for Δθ in equation (IV) to find |Δv|.

    |Δv|=(2.98×104m/s)(0.0172rad)=513.5m/s

Therefore, the change in Earths velocity is 513.5m/s_ toward the center of Sun.

(c)

To determine

The average acceleration in one day.

(c)

Expert Solution
Check Mark

Answer to Problem 92P

The average acceleration in one day is 0.00594m/s2_.

Explanation of Solution

Write the expression for the average acceleration during 1 day,

    a=vt        (V)

Conclusion:

Substitute 513.5m/s for v and 86400s for t in equation (V) to find a.

    a=513.5m/s86400s=0.00594m/s2

Therefore, The average acceleration in one day is 0.00594m/s2_.

(d)

To determine

The radial acceleration of earth.

(d)

Expert Solution
Check Mark

Answer to Problem 92P

The radial acceleration of earth is 0.00594m/s2_.

Explanation of Solution

Write the expression for the radial acceleration of the Earth.

    aR=v2r        (VI)

Conclusion:

Substitute 513.5m/s for v and 1.5×1011m for r in equation (V) to find aR.

    a=(513.5m/s)21.5×1011m=0.00594m/s2

Therefore, The average acceleration in one day is 0.00594m/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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