(a) The change in volume of the sphere

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

To determine

(b)

The radius of the sphere.

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Fresnel lens: You would like to design a 25 mm diameter blazed Fresnel zone plate with a first-order power of +1.5 diopters. What is the lithography requirement (resolution required) for making this lens that is designed for 550 nm? Express your answer in units of μm to one decimal point. Fresnel lens: What would the power of the first diffracted order of this lens be at wavelength of 400 nm? Express your answer in diopters to one decimal point. Eye: A person with myopic eyes has a far point of 15 cm. What power contact lenses does she need to correct her version to a standard far point at infinity? Give your answer in diopter to one decimal point.

Paraxial design of a field flattener. Imagine your optical system has Petzal curvature of the field with radius p. In Module 1 of Course 1, a homework problem asked you to derive the paraxial focus shift along the axis when a slab of glass was inserted in a converging cone of rays. Find or re-derive that result, then use it to calculate the paraxial radius of curvature of a field flattener of refractive index n that will correct the observed Petzval. Assume that the side of the flattener facing the image plane is plano. What is the required radius of the plano-convex field flattener? (p written as rho )

3.37(a) Five free electrons exist in a three-dimensional infinite potential well with all three widths equal to \( a = 12 \, \text{Å} \). Determine the Fermi energy level at \( T = 0 \, \text{K} \). (b) Repeat part (a) for 13 electrons. Book: Semiconductor Physics and Devices 4th ed, NeamanChapter-3Please expert answer only. don't give gpt-generated answers, & please clear the concept of quantum states for determining nx, ny, nz to determine E, as I don't have much idea about that topic.

Answer 2

Question

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

To determine

(b)

The radius of the sphere.

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

To determine

(b)

The radius of the sphere.

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

To determine

(b)

The radius of the sphere.

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

To determine

(b)

The radius of the sphere.

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Answer 6

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

Answer 7

Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Answer 8

Expert Solution

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×10−5 m3.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Bulk Modulus of copper B=140×109 Nm−2

Surface Area of sphere A=4πr02=4π×(5.00× 10 −2m)2=3.14×10−2m2

Let the change in volume be ΔV

Applied force F=2.00×108 N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

ΔV=FB×V0AΔV=2.00× 108 N140× 109 Nm −2×5.24× 10 −4m33.14× 10 −2m2ΔV=2.38×10−5 m3

Hence, the change in volume of the sphere is 2.38×10−5 m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×10−5 m3.

Answer 13

To determine

(b)

The radius of the sphere.

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Answer 14

To determine

(b)

The radius of the sphere.

Answer 15

Expert Solution

Answer to Problem 52QAP

The final radius of the sphere is 5.08 cm.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Original radius of sphere r0=5.00 cm=5.00×10−2m

Original volume of sphere V0=43πr03=43π(5.00× 10 −2m)3=5.24×10−4m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×10−5 m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

Vf=ΔV+V0Vf=2.38×10−5 m3+5.24×10−4m3Vf=0.238×10−4 m3+5.24×10−4m3Vf=5.48×10−4 m3

But final volume of sphere Vf=43πrf3=5.48×10−4 m3

So, final radius of the sphere

rf=3 4π×5.48× 10 −4 m33rf=3 4π×5.48× 10 −4 m33rf=130.8× 10 −6 m33rf=5.08×10−2 mrf=5.08 cm

Hence, the final radius of the sphere is 5.08 cm.

Conclusion:

Thus, the final radius of the sphere is 5.08 cm.

Answer 20

Ch. 9 - Prob. 1QAP Ch. 9 - Prob. 2QAP Ch. 9 - Prob. 3QAP Ch. 9 - Prob. 4QAP Ch. 9 - Prob. 5QAP Ch. 9 - Prob. 6QAP Ch. 9 - Prob. 7QAP Ch. 9 - Prob. 8QAP Ch. 9 - Prob. 9QAP Ch. 9 - Prob. 10QAP

(a) The change in volume of the sphere

Concept explainers

Videos

(a)

The change in volume of the sphere

Answer to Problem 52QAP

Explanation of Solution

(b)

The radius of the sphere.

Answer to Problem 52QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions

(a) The change in volume of the sphere

Concept explainers

Videos

(a) The change in volume of the sphere

Answer to Problem 52QAP

Explanation of Solution

(b) The radius of the sphere.

Answer to Problem 52QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions

(a)

The change in volume of the sphere

(b)

The radius of the sphere.