General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 20, Problem 47E

(a)

To determine

The number of bits per centimeter stored on a track.

(a)

Expert Solution
Check Mark

Answer to Problem 47E

The number of bits stored per centimeter is 1592bitscm-1.

Explanation of Solution

Write the expression for the number of bits stored on a track.

    n=Ul        (1)

Here, n is the number of bits, U is the transfer rate and l is the length covered by the track.

Write the expression for the length of the track.

    l=ωr        (2)

Here, ω is the spin rate and r is the radius of the track.

Conclusion:

Substitute 300revmin-1 for ω and 5cm for r in equation (2).

    l=(300revmin-1(2πrad1rev)(1min60s))(5cm)=50πcms-1

Substitute 250kilobitss-1 for U and 50πcms1 for l in equation (1).

    n=250kilobitss-1(103bits1kilobits)50πcms-1=1592bitscm-1

Thus, the number of bits stored per centimeter is 1592bitscm-1.

(a)

To determine

The number of bits stored per unit length of the track.

(a)

Expert Solution
Check Mark

Answer to Problem 47E

The number of bits stored per centimeter is 4406bitscm-1.

Explanation of Solution

Write the expression for the number of bits stored on a track.

    n=Ul

Write the expression for the length of the track.

    l=ωr

Conclusion:

Substitute 3633revmin-1 for ω and 5cm for r in equation (2).

    l=(3633revmin-1(2πrad1rev)(1min60s))(10.5cm)=3994.7cms-1

Substitute 17.6megabitss-1 for U and 3994.7cms-1 for l in equation (1).

    n=17.6megabitss-1(106bits1megabits)3994.7cms-1=4406bitscm-1

Thus, the number of bits stored per centimeter is 4406bitscm-1.

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Chapter 20 Solutions

General Physics, 2nd Edition

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