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 84P

(a)

To determine

The radial acceleration of the material in the centrifugal conditions.

(a)

Expert Solution
Check Mark

Answer to Problem 84P

The radial acceleration of the material in the centrifugal conditions is 89.36g_.

Explanation of Solution

Write the expression for the speed of the material,

    v=2πrf        (I)

Here, v is the speed of the material, r is the radius of the rotation and f is the frequency.

Write the expression for the radial acceleration of the material in the centrifugal conditions,

    a=v2r        (II)

Here, a is the radial acceleration of the material in the centrifugal conditions.

Conclusion:

Substitute 0.080m for r and 1000rev/min for f in (I) to find v,

    v=2π(0.080m)(1000revmin)×(1min60s)=8.37m/s

Substitute 8.37m/s for v and 0.080m for r in (II) to find a ,

    a=(8.37m/s)20.080m=875.71m/s2

In terms of g,

    a=875.71m/s29.8m/s2g=89.36g

Therefore, the radial acceleration of the material in the centrifugal conditions is 89.36g_.

(b)

To determine

The net force on the red blood cell.

(b)

Expert Solution
Check Mark

Answer to Problem 84P

The net force on the red blood cell is 7.88×1011N_.

Explanation of Solution

Write the expression for the net force on the red blood cell,

    F=ma        (III)

Here, F is the net force on the red blood cell, a is the acceleration and m is the mass of the red blood cell.

Conclusion:

Substitute 9.0×1014kg for m and 875.71m/s2 for a in (III) to find F,

    F=(9.0×1014kg)(875.71m/s2)=7.88×1011N

Therefore, the net force on the red blood cell is 7.88×1011N_.

(c)

To determine

The net force on the virus particle.

(c)

Expert Solution
Check Mark

Answer to Problem 84P

The net force on the virus particle is 4.38×1018N_.

Explanation of Solution

Write the expression for the net force on the virus particle,

    F=ma        (IV)

Here, F is the net force on the virus particle, a is the acceleration and m  is the mass of the virus particle.

Conclusion:

Substitute 5.0×1021kg for m and 875.71m/s2 for a in (IV) to find F,

    F=(5.0×1021kg)(875.71m/s2)=4.38×1018N

Therefore, the net force on the virus particle is 4.38×1018N_.

(d)

To determine

The radial acceleration of the material in the ultra-centrifuge conditions.

(d)

Expert Solution
Check Mark

Answer to Problem 84P

The radial acceleration of the material in the ultra-centrifuge conditions is 503040.8g_.

Explanation of Solution

Write the expression for the speed of the material,

    v=2πrf        (V)

Here, v is the speed of the material, r is the radius of the rotation and f is the frequency.

Write the expression for the radial acceleration of the material in the ultra-centrifuge conditions,

    a=v2r        (VI)

Here, a is the radial acceleration of the material in the ultra-centrifuge conditions.

Conclusion:

Substitute 0.080m for r and 75000rev/min for f in (V) to find v,

    v=2π(0.080m)(75000revmin)×(1min60s)=628m/s

Substitute 628m/s for v and 0.080m for r in (VI) to find a ,

    a=(628m/s)20.080m=4929800m/s2=4.93×106m/s2

In terms of g ,

    a=4929800m/s29.8m/s2g=503040.8g

Therefore, the radial acceleration of the material in the ultracentrifuge conditions is 503040.8g_.

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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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