EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
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Chapter 3, Problem 60PQ

(a)

To determine

The vector A+B

(a)

Expert Solution
Check Mark

Answer to Problem 60PQ

The vector A+B is 6i^2j^.

Explanation of Solution

Write an expression for the vector A.

    A=Axi^+Ayj^ (I)

Here, A is the vector, Ax is the horizontal component of the vector and Ay is the vertical component of the vector A.

Write an expression for the vector B.

    B=Bxi^+Byj^ (II)

Here, B is the vector, Bx is the horizontal component of the vector and By is the vertical component of the vector B.

Write an expression for A+B.

    A+B=(Ax+Bx)i^+(Ay+By)j^ (III)

Here, A+B is the sum of the vectors.

Conclusion:

The vectors are A=4i+3j^ and B=2i^5j^.

Substitute 4 for Ax, 3 for Ay, 2 for Bx and 5 for By in equation (III) to find A+B.

    A+B=(4+2)i^+(3+5)j^=(6)i^+(2)j^

Thus, the vector A+B is 6i^2j^.

(b)

To determine

The vector AB

(b)

Expert Solution
Check Mark

Answer to Problem 60PQ

The vector AB is 2i^+8j^.

Explanation of Solution

Write an expression for AB.

    AB=(AxBx)i^+(AyBy)j^ (IV)

Here, AB is the difference of the vectors.

Conclusion:

Substitute 4 for Ax, 3 for Ay, 2 for Bx and 5 for By in equation (IV) to find AB.

    A+B=(4(2))i^+(3+(5))j^=(2)i^+(8)j^

Thus, the vector A+B is 2i^+8j^.

(c)

To determine

The magnitude of A+B.

(c)

Expert Solution
Check Mark

Answer to Problem 60PQ

The magnitude of A+B is 6.3.

Explanation of Solution

Write an expression for magnitude of A+B.

    |A+B|=(A+B)x2+(A+B)y2 (V)

Here, |A+B| is the magnitude of A+B, (A+B)x is the x component of A+B and (A+B)y is the y component of A+B

Conclusion:

Substitute 6 for (A+B)x and 2 for (A+B)y in equation (V) to find |A+B|.

    |A+B|=(6)2+(2)2=36+4=6.3

Thus, the magnitude of A+B is 6.3.

(d)

To determine

The magnitude of AB.

(d)

Expert Solution
Check Mark

Answer to Problem 60PQ

The magnitude of AB is 8.2.

Explanation of Solution

Write an expression for magnitude of AB.

    |AB|=(AB)x2+(AB)y2 (VI)

Here, |AB| is the magnitude of AB, (AB)x is the x component of AB and (AB)y is the y component of AB

Conclusion:

Substitute 2 for (AB)x and 8 for (AB)y in equation (VI) to find |AB|.

    |AB|=(2)2+(8)2=4+64=8.2

Thus, the magnitude of AB is 8.2.

(e)

To determine

The direction of A+B and AB.

(e)

Expert Solution
Check Mark

Answer to Problem 60PQ

The direction of A+B is 18.4°below the -x axis and direction of AB is 76.0°above the -x axis.

Explanation of Solution

Write an expression for direction of A+B.

    θA+B=tan1(A+B)y(A+B)x (VII)

Here, θA+B is the direction of A+B.

Write an expression for direction of AB.

    θAB=tan1(AB)y(AB)x (VIII)

Here, θAB is the direction of AB.

Conclusion:

Substitute 6 for (A+B)x and 2 for (A+B)y in equation (VII) to find θA+B.

    θA+B=tan1(26)=tan1(13)=18.4°below the -x axis

Substitute 2 for (AB)x and 8 for (AB)y in equation (VIII) to find θAB.

    θAB=tan1(82)=tan1(4)=76.0°above the -x axis

Thus, the direction of A+B is 18.4°below the -x axis and direction of AB is 76.0°above the -x axis.

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Chapter 3 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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