Using the property indicated in Prob. 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w 0 cos ( πx / L ), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable. PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w ( x ) is defined by the differential equation d 2 y / dx 2 = w ( x )/ T 0 , where T 0 is the tension at the lowest point.

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Answer 1

Textbook Question

Chapter 7.4, Problem 7.125P

Using the property indicated in Prob. 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w₀ cos (πx/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable.

PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d²y/dx² = w(x)/T₀, where T₀ is the tension at the lowest point.

Expert Solution & Answer

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

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Answer 2

Textbook Question

Chapter 7.4, Problem 7.125P

Using the property indicated in Prob. 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w₀ cos (πx/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable.

PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d²y/dx² = w(x)/T₀, where T₀ is the tension at the lowest point.

Expert Solution & Answer

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

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Answer 3

Textbook Question

Chapter 7.4, Problem 7.125P

Using the property indicated in Prob. 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w₀ cos (πx/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable.

PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d²y/dx² = w(x)/T₀, where T₀ is the tension at the lowest point.

Expert Solution & Answer

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

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Answer 4

Textbook Question

Chapter 7.4, Problem 7.125P

Using the property indicated in Prob. 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w₀ cos (πx/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable.

PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d²y/dx² = w(x)/T₀, where T₀ is the tension at the lowest point.

Expert Solution & Answer

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Answer 8

Expert Solution & Answer

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer 12

Answer to Problem 7.125P

The curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Answer 13

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Vector Mechanics for Engineers: Statics, Chapter 7.4, Problem 7.125P

Write the expression for the load distributed.

w(x)=w0cosπxL

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

d2ydx2=w(x)T0

Substitute w0cosπxL for w(x) in the above equation to rewrite.

d2ydx2=w0cosπxLT0

Integrate both side of the above equation and apply the condition dydx|x=0=0.

dydx=∫w0cosπxLT0dx=w0LT0πsinπxL

Integrate the above equation and apply the condition y|x=0=0.

y=∫(w0LT0π)sinπxLdx=(w0L2T0π2)(−cosπxL)+C

Here C is the integration constant.

Apply the condition y|x=0=0 to calculate C.

0=(w0L2T0π2)(−cos0)+CC=w0L2T0π2

Substitute w0L2T0π2 for C in the equation for y to rewrite.

y=(w0L2T0π2)(−cosπxL)+w0L2T0π2=(w0L2T0π2)(1−cosπxL)

At x=L2 the tension on the cable might have minimum value.

Substitute x=L2 in the above equation for y to calculate the minimum tension.

y=(w0L2T0π2)(1−cosπL(L2))=(w0L2T0π2)(1−cosπ2)=w0L2T0π2

The value of y at x=L2 is same as the sag h.

Rewrite the above equation in terms of h.

h=w0L2T0π2T0=w0L2hπ2

Here T0 is the minimum tension on the cable.

Therefore the minimum tension on the cable is w0L2T0π2.

To find the maximum tension on the cable the slope of the above equation for y at x=L2 which is equal to the ratio TByT0

Here TBy is the y-component of the tension at A.

dydx=TByT0=ddx((w0L2T0π2)(1−cosπxL))=(w0L2T0π2)(πL)sinπxLTByT0=(w0LT0π)sinπxL

Substitute x=L2 in the above equation to calculate TBy.

TByT0=(w0LT0π)sinπL(L2)=(w0LT0π)sinπ2=w0LT0π

Rewrite the above equation in terms of TBy.

TBy=(w0LT0π)T0=w0Lπ

Write the expression to calculate the maximum tension on the cable.

Tmax=TBy2+T02

Here Tmax is the maximum tension in the cable.

Substitute w0Lπ for TBy and w0L2hπ2 for T0 in the above equation to calculate Tmax.

Tmax=(w0Lπ)2+(w0L2hπ2)2=(w0Lπ)1+(Lπh)2

Conclusion:

Thus, the curve represented by the cable is (w0L2T0π2)(1−cosπxL)_, the minimum tension on the cable is w0L2T0π2_ and maximum tension on the cable is (w0Lπ)1+(Lπh)2_.

Answer 14

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Answer to Problem 7.125P

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