PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 1, Problem 72P

(a)

To determine

The diameter of a viroid in m .

(a)

Expert Solution
Check Mark

Answer to Problem 72P

The diameter of a viroid in m is 3.3×108 m .

Explanation of Solution

The viroid can be considered as circular loop containing 300 bases as beads. It is given that the distance from one base to the next is 0.35 nm .

Find the circumference of the viroid.

C=300×0.35 nm(1 m109 nm)=300×(0.35×109 m)=1.05×107 m

Here, C is the circumference of the circle

Write the equation for the circumference of a circle.

C=πd

Here, d is the diameter of the circle

Rewrite the above equation for d .

d=Cπ (I)

Conclusion:

Substitute 1.05×107 m for C in equation (I) to find the value of d .

d=1.05×107 mπ=3.3×108 m

Therefore, the diameter of a viroid in m is 3.3×108 m .

(b)

To determine

The diameter of a viroid in μm .

(b)

Expert Solution
Check Mark

Answer to Problem 72P

The diameter of a viroid in μm is 3.3×102 μm .

Explanation of Solution

Write the expression to convert distance in m into μm .

d(μm)=d(m)106 μm1 m (II)

Here, d(μm) is the quantity in μm and d(m) is the quantity in m .

Conclusion:

Substitute 3.3×108 m for d(m) in equation (II) to find d(μm) .

d(μm)=3.3×108 m106 μm1 m=3.3×102 m

Therefore, the diameter of a viroid in μm is 3.3×102 μm .

(c)

To determine

The diameter of a viroid in in .

(c)

Expert Solution
Check Mark

Answer to Problem 72P

The diameter of a viroid in in is 1.3×106 in .

Explanation of Solution

Write the expression to convert distance in m into in .

d(in)=d(m)39.37 in1 m (III)

Here, d(in) is the quantity in in

Conclusion:

Substitute 3.3×108 m for d(m) in equation (III) to find d(in) .

d(μm)=3.3×108 m39.37 in1 m=1.3×106 in

Therefore, the diameter of a viroid in in is 1.3×106 in .

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Chapter 1 Solutions

PHYSICS

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