Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
Question
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Chapter 3, Problem 3.23P

(a)

To determine

The addition of two vector A and B .

(a)

Expert Solution
Check Mark

Answer to Problem 3.23P

The addition of two vector A and B is 2i^6j^ .

Explanation of Solution

Given info: The value of two vector A and B is 3i^2j^ and i^4j^ respectively.

Write the expression for addition of two vector A and B .

R1=A+B (1)

Substitute 3i^2j^ for A and i^4j^ for B in equation (1) to get value of two vector A and B .

R1=(3i^2j^)+(i^4j^)=(3i^i^)+(2j^4j^)=(2i^)+(6j^)=2i^6j^

Conclusion:

Therefore, the addition of two vector A and B is 2i^6j^ .

(b)

To determine

The subtraction of two vector A and B .

(b)

Expert Solution
Check Mark

Answer to Problem 3.23P

The subtraction of two vector A and B is 4i^+2j^ .

Explanation of Solution

Given info: The value of two vector A and B is 3i^2j^ and i^4j^ respectively.

Write the expression for subtraction of two vector A and B .

R2=AB (2)

Substitute 3i^2j^ for A and i^4j^ for B in equation (2) to get value of two vector A and B .

R2=(3i^2j^)(i^4j^)=(3i^+i^)+(2j^+4j^)=(4i^)+(2j^)=4i^+2j^

Conclusion:

Therefore, the subtraction of two vector A and B is 4i^+2j^ .

(c)

To determine

The magnitude of addition of two vector A and B .

(c)

Expert Solution
Check Mark

Answer to Problem 3.23P

The magnitude of two vector A and B is 6.32 .

Explanation of Solution

Given info: The value of two vector A and B is 3i^2j^ and i^4j^ respectively.

Write the expression for addition of two vector A and B .

R1=A+B (3)

Substitute 3i^2j^ for A and i^4j^ for B in equation (3) to get value of two vector A and B .

R1=(3i^2j^)+(i^4j^)=(3i^i^)+(2j^4j^)=(2i^)+(6j^)=2i^6j^

Write the expression for the magnitude of two vectors A and B is,

|R1|=(R1)x2+(R1)y2 (4)

Here,

(R1)x is x -axis coordinate.

(R1)y is y -axis coordinate.

Substitute 2 for (R1)x and 6 for (R1)y in equation (4) to get magnitude of two

Vector A and B .

|R1|=(2)2+(6)2=4+36=40=6.32

Conclusion:

Therefore, the magnitude of two vector A and B is 6.32 .

(d)

To determine

The magnitude of subtraction of two vector A and B .

(d)

Expert Solution
Check Mark

Answer to Problem 3.23P

The magnitude of two vector A and B is 4.47 .

Explanation of Solution

Given info: The value of two vector A and B is 3i^2j^ and i^4j^ respectively

Write the expression for subtraction of two vector A and B .

R2=AB (5)

Substitute 3i^2j^ for A and i^4j^ for B in equation (5) to get value of two vector A and B .

R2=(3i^2j^)(i^4j^)=(3i^+i^)+(2j^+4j^)=(4i^)+(2j^)=4i^+2j^

Write the expression for the magnitude of two vector A and B is,

|R2|=(R2)x2+(R2)y2 (6)

Here,

(R2)x is x -axis coordinate.

(R2)y is y -axis coordinate.

Substitute 4 for (R2)x and 2 for (R2)y in equation (6) to get magnitude of two

Vector A and B .

|R2|=(4)2+(2)2=16+4=20=4.47

Conclusion:

Therefore, the magnitude of two vector A and B is 4.47 .

(e)

To determine

The direction of vector |R1| and |R2| .

(e)

Expert Solution
Check Mark

Answer to Problem 3.23P

The direction of vector |R1| is 288° counter clockwise with positive x -axis and the direction of vector |R2| is 26.6° counter clockwise with positive x -axis

Explanation of Solution

Given info: The value of vector |R1| and |R2| is 2i^6j^ and 4i^+2j^ respectively.

Write the expression for direction of |R1| .

tanθ|R1|=(|R1|)y(|R1|)x=tan1((|R1|)y(|R1|)x) (7)

Here,

(|R1|)x is the horizontal component of vector |R1| .

(|R1|)y is the vertical component of vector |R1| .

Substitute 6 for (|R1|)y and 2 for (|R1|)x in equation (7).

θ|R1|=tan1(62)=tan1(3.0)=71.56°

The angle made by the vector with the positive axis is,

θ|R1|=360°+|θ|R1||

Substitute 71.56° for |θ|R1|| in above equation to get the angle made by the vector with the positive axis.

θ|R1|=360°71.56°=288°

Write the expression for direction of |R2| .

tanθ|R2|=(|R2|)y(|R2|)x=tan1((|R2|)y(|R2|)x) (8)

Here,

(|R2|)x is the horizontal component of vector |R2| .

|R2| is the vertical component of vector |R2| .

Substitute 2 for (|R2|)x and 4 for |R2| in equation (8).

θ|R2|=tan1(24)=tan1(0.5)=26.6°

Conclusion:

Therefore, the direction of vector |R1| is 288° counter clockwise with positive x -axis and the direction of vector |R2| is 26.6° counter clockwise with positive x -axis.

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Chapter 3 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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