A → × B →

Question

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Answer 1

Question

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

(b)

To determine

To Find: A→×B→

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

(c)

To determine

To Find: A→×B→

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

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Students have asked these similar questions

A sinusoidal wave is propagating along a stretched string that lies along the x-axis. The displacement of the string as a function of time is graphed in (Figure 1) for particles at x = 0 and at x = 0.0900 m. You are told that the two points x = 0 and x = 0.0900 m are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength. If instead the wave is moving in the -x-direction, determine the wavelength. Please show all steps

You are designing a two-string instrument with metal strings 35.0 cm long, as shown in (Figure 1). Both strings are under the same tension. String S1 has a mass of 8.30 g and produces the note middle C (frequency 262 Hz ) in its fundamental mode. What should be the tension in the string? What should be the mass of string S2 so that it will produce A-sharp (frequency 466 Hz ) as its fundamental? To extend the range of your instrument, you include a fret located just under the strings but not normally touching them. How far from the upper end should you put this fret so that when you press S1 tightly against it, this string will produce C-sharp (frequency 277 Hz ) in its fundamental? That is, what is x in the figure? If you press S2 against the fret, what frequency of sound will it produce in its fundamental?

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Answer 2

Question

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

(b)

To determine

To Find: A→×B→

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

(c)

To determine

To Find: A→×B→

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

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Answer 3

Question

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

(b)

To determine

To Find: A→×B→

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

(c)

To determine

To Find: A→×B→

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

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Answer 4

Question

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

(b)

To determine

To Find: A→×B→

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

(c)

To determine

To Find: A→×B→

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

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Answer 5

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

(b)

To determine

To Find: A→×B→

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

(c)

To determine

To Find: A→×B→

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

Answer 6

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

Answer 7

Chapter 10, Problem 29P

(a)

To determine

To find: A→×B→

Answer 8

(a)

Expert Solution

Answer to Problem 29P

A→×B→=24 k^

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 j^

Formula used:

The property of vector cross product:

i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 j^)=4×6(i^×i^)+4×6(i^×j^)=24×0+24 k^=24 k^

Conclusion:

A→×B→=24 k^

Answer 13

(b)

To determine

To Find: A→×B→

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

Answer 14

(b)

To determine

To Find: A→×B→

Answer 15

(b)

Expert Solution

Answer to Problem 29P

A→×B→=−24 j^

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

A→=4 i^B→=6 i^+6 k^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=4 i^×(6 i^+6 k^)=4×6(i^×i^)+4×6(i^×k^)=24×0+24(- j^)=−24 j^

Conclusion:

A→×B→=−24 j^

Answer 20

(c)

To determine

To Find: A→×B→

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

Answer 21

(c)

To determine

To Find: A→×B→

Answer 22

(c)

Expert Solution

Answer to Problem 29P

A→×B→=−5k^

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

A→=2 i^+3 j^B→=3 i^+2 j^

Formula used:

The property of vector cross product:

· i^ ×j^=k^, j^ ×k^=i^, k^ ×i^=j^· j^ ×i^=−k^, k^ ×j^=−i^, i^ ×k^=−j^· i^ ×i^=0, j^ ×j^=0, k^ ×k^=0

Calculation:

A→×B→=(2 i^+3 j^)×(3 i^+2 j^)=2×3(i^×i^)+2×2(i^×j^)+3×3(j^×i^)+3×2(j^×j^)=6×0+4k^+9(−k^)+6×0=4k^−9k^A→×B→=−5k^

Conclusion:

A→×B→=−5k^

Answer 27

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

A → × B →

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To find: A→×B→

Answer to Problem 29P

Explanation of Solution

To Find: A→×B→

Answer to Problem 29P

Explanation of Solution

To Find: A→×B→

Answer to Problem 29P

Explanation of Solution

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