Physics for Scientists and Engineers, Volume 1
Physics for Scientists and Engineers, Volume 1
9th Edition
ISBN: 9781133954156
Author: Raymond A. Serway
Publisher: CENGAGE L
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Chapter 1, Problem 1.51P

Review. A student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. He cuts out various shapes in various sizes, calculates their areas, measures their masses, and prepares the graph of Figure PI.51. (a) Consider the fourth experimental point from the top. How far is it from the best-lit straight line? Express your answer as a difference in vertical-axis coordinate, (b) Express your answer as a percentage, (c) Calculate the slope of the line, (d) State what the graph demonstrates, referring to the shape of the graph and the results of parts (b) and (c). (e) Describe whether this result should be expected theoretically, (f) Describe the physical meaning of the slope.

Chapter 1, Problem 1.51P, Review. A student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive

(a)

Expert Solution
Check Mark
To determine

The distance of the fourth experimental point from the top from the best –fit straight line.

Answer to Problem 1.51P

The distance of the fourth experimental point from the top from the best fit straight line is 0.015g .

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

Consider the following figure.

Physics for Scientists and Engineers, Volume 1, Chapter 1, Problem 1.51P

Figure (1)

Figure indicates the graph plotted area of pieces versus mass of the pieces of the paper.

The fourth experimental point from the top is a circle. It lies slightly above the best fit line.

From figure (1), the vertical coordinate for the forth experiment is 0.2025g and the vertical coordinate for the point on best fit line at which it touches the best fit line is 0.1875g .

So the difference in the vertical axis coordinate is,

D=P1P2

Here,

P1 is the vertical coordinate of the fourth experiment.

P2 is the vertical coordinate of the point on best fit line at which it touches the best fit line.

Substitute 0.2025g for P1 and 0.1875g for P2 in above expression.

d=0.2025g0.1875g=0.015g

Conclusion:

Therefore the distance of the fourth experimental point from the top from the best fit straight line is 0.015g .

(b)

Expert Solution
Check Mark
To determine

The answer in the form of percentage.

Answer to Problem 1.51P

The answer in the form of percentage is 8% .

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The expression for the percentage is,

P=DP1×100

Substitute 0.015g for D and 0.1875g for P1 in above expression.

P=0.015g0.1875g×100=8%

Conclusion:

Therefore the answer in the form of percentage is 8% .

(c)

Expert Solution
Check Mark
To determine

The slope of the line.

Answer to Problem 1.51P

The slope of the line is 5.2g/m2 .

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The expression for the slope is,

S=y2y1x2x1

Here,

y1 is the first vertical coordinate.

y2 is the second vertical coordinate.

x1 is the first vertical coordinate.

x2 is the second vertical coordinate.

Substitute 0.2g for y2 , 0.1g for y1 , 392cm2 for x2 and 200cm2 for x1 in above expression.

S=0.2g0.1g392cm2200cm2=5.2×104g/cm2×10000cm21m2=5.2g/m2

Conclusion:

Therefore the slope of the line is 5.2g/m2 .

(d)

Expert Solution
Check Mark
To determine

The demonstration from the graph referring to the shape of the graph and the results of part (b) and (c).

Answer to Problem 1.51P

The graph demonstrates that the mass of the cutout is proportional to its area for the shape cuts from this copy paper and the proportionality constant is 5.2g/m2±8% where the uncertainty is estimated.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The graph given figure (1) demonstrates that the mass of the cutout in each shape is proportional to its area for the shape cuts from this copy paper and the proportionality constant is 5.2g/m2±8% where the 8% uncertainty is estimated in the aerial mss density of each shape.

The value of the slope from part (c) of the question and from the part (b) question the percentage uncertainty is 8% .

Conclusion:

Therefore the graph demonstrates that the mass of the cutout is proportional to its area for the shape cuts from this copy paper and the proportionality constant is 5.2g/m2±8% where the uncertainty is estimated.

(e)

Expert Solution
Check Mark
To determine

Whether this result should be expected theoretically or not.

Answer to Problem 1.51P

This result is to be expected theoretically if the paper has thickness and density that are uniform within the experimental uncertainty.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The graph given figure (1) demonstrates that the mass of the cutout in each shape is proportional to its area for the shape cuts from this copy paper and the proportionality constant is 5.2g/m2±8% where the 8% uncertainty is estimated in the aerial mss density of each shape.

The value of the slope from part (c) of the question is 5.2g/m2 and from the part (b) question the percentage uncertainty is 8% .

Thus this result is expected theoretically when the object is having same aerial mass density and the paper has thickness and density that are uniform within the experimental uncertainty.

Conclusion:

Therefore this result is to be expected theoretically if the paper has thickness and density that are uniform within the experimental uncertainty.

(f)

Expert Solution
Check Mark
To determine

The physical meaning of the slope.

Answer to Problem 1.51P

The physical meaning of the slope is the aerial density of the paper that represents the mass per unit area.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The graph given figure (1) demonstrates that the mass of the cutout in each shape is proportional to its area for the shape cuts from this copy paper and the proportionality constant is 5.2g/m2±8% where the 8% uncertainty is estimated in the aerial mss density of each shape.

The value of the slope from part (c) of the question is 5.2g/m2 and from the part (b) question the percentage uncertainty is 8% .

The physical meaning of the slope is the aerial density of the paper that represent the mass per unit area.

Conclusion:

Therefore the physical meaning of the slope is the aerial density of the paper that represents the mass per unit area.

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Physics for Scientists and Engineers, Volume 1

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