PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES
PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES
9th Edition
ISBN: 9780357001417
Author: SERWAY
Publisher: CENGAGE L
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Chapter 6, Problem 49AP

(a)

To determine

The slope of the straight line of the graph.

(a)

Expert Solution
Check Mark

Answer to Problem 49AP

The slope of the straight line of the graph is 0.0162kg/m.

Explanation of Solution

Write the expression for the slope of the graph line.

    s=(y2y1)(x2x1)                                                                                   (I)

Here, s is the slope of the graph line, x1 is the initial point on the resistive force, x2 is the final point on the resistive force, y1 is initial point on the square of the terminal speed and y2 is initial point on the square of the terminal speed.

Conclusion:

Substitute 0N for y1, 0.160N for y2, 0(m/s)2 for x1 and 9.9(m/s)2 for x2 in (I) to find s.

    s=(0.160N0N)(9.9(m/s)20(m/s)2)=0.0162kg/m

Therefore, the slope of the straight line of the graph is 0.0162kg/m.

(b)

To determine

The theoretical slope of a graph of resistive force versus squared speed.

(b)

Expert Solution
Check Mark

Answer to Problem 49AP

The theoretical slope of a graph of resistive force versus squared speed is 12DρA.

Explanation of Solution

Write the expression for the resistive force of the object.

    R=12DρAv2                                                                                      (II)

Here, R is the resistive force of the object, D is a drag coefficient, ρ is a density of air, A is the cross-sectional area of the moving object and v is speed of the moving object.

Write the expression for the theoretical slope of a graph.

    s=Rv2                                                                                                  (III)

Here, s is the theoretical slope of a graph and R is the resistive force of the object.

Conclusion:

Substitute 12DρAv2 for R in (II) to find s.

    s=12DρAv2v2=12DρA

Therefore, the theoretical slope of a graph of resistive force versus squared speed is 12DρA.

(c)

To determine

The drag coefficient of the filters.

(c)

Expert Solution
Check Mark

Answer to Problem 49AP

The drag coefficient of the filters is 0.778.

Explanation of Solution

Write the expression for the cross-sectional area of the filters.

    A=πr2                                                                                           (IV)

Here, r is the radius of the filters.

From part (b),

Write the expression for the theoretical slope of a graph of resistive force versus squared speed.

    s=12DρA                                                                                           (V)

Here, s is the theoretical slope of a graph of resistive force versus squared speed D is a drag coefficient, ρ is a density of air and A is the cross-sectional area of the moving object.

Rewrite the above expression for the drag coefficient of filters.

    D=2sρA                                                                                            (VI)

Substitute πr2 for A in the above equation.

    D=2sρπr2                                                                                         (VII)

Conclusion:

Substitute 0.0162kg/m for s, 1.20kg/m3 for ρ and 0.105m for r in (VII) to find D.

    D=2(0.0162kg/m)(1.20kg/m3)π(0.105m)2=0.778

Therefore, the drag coefficient of the filters is 0.778 .

(d)

To determine

The percentage of the scatter.

(d)

Expert Solution
Check Mark

Answer to Problem 49AP

The percentage of the scatter is 1.5%.

Explanation of Solution

In this case the eighth point data is at force.

Write the expression for the magnitude of the resistive force at the eighth point.

    Rx=8mg                                                                                        (VIII)

Here, Rx is the magnitude of the resistive force at the eighth point, m is the mass of the object and g is the gravitational due to gravity.

Write the expression for the vertical resistive force.

    Ry=svx2                                                                                             (IX)

Here, Ry is the vertical resistive force vx is the horizontal speed.

Write the expression for the percentage of the scatter.

    S%=(RyRxRx)×100%                                                                      (X)

Here, S% is the percentage of the scatter,

Conclusion:

Substitute 1.64×103kg for m and 9.8m/s2 for g in (VIII) to find Rx.

    Rx=8(1.64×103kg)(9.8m/s2)=0.129N

Substitute 0.0162kg/m for s, 2.80m/s for vx in (IX) to find Ry.

    Ry=(0.0162kg/m)(2.80m/s)2=0.127N

Substitute 0.129N for Rx and 0.127N for Ry in (X) to find S%.

    S%=(0.129N0.127N0.127N)×100%=1.5%

Therefore, the percentage of the scatter is 1.5%.

(e)

To determine

The state of the graph demonstrates and compares it with theoretical prediction.

(e)

Expert Solution
Check Mark

Answer to Problem 49AP

The state of the graph demonstrates that the resistive force is proportional to the squared speed within an estimated as 2%.

Explanation of Solution

In this case, the coffee filters falling with a terminal speed.

A graph of air resistance force as a function of speed squared.

It demonstrated that the force of the falling coffee filters is proportional to the squared speed.

It is with an estimated value of 2% . It is also agree the theoretical equation.

Conclusion:

The value of the slope of the graph indicates that the drag coefficient for the coffee filters is about 0.78±2%.

Therefore, the state of the graph demonstrates that the resistive force is proportional to the squared speed within an estimated as 2%.

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Chapter 6 Solutions

PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES

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