The value of A from scalar quantity 4 ( 7 + 3 ) .

Question

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Answer 1

Question

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

(c)

To determine

The proper way to get the answer and convince the student.

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

(e)

To determine

The way to get the answer.

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

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Answer 2

Question

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

(c)

To determine

The proper way to get the answer and convince the student.

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

(e)

To determine

The way to get the answer.

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

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Answer 3

Question

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

(c)

To determine

The proper way to get the answer and convince the student.

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

(e)

To determine

The way to get the answer.

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

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Answer 4

Question

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

(c)

To determine

The proper way to get the answer and convince the student.

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

(e)

To determine

The way to get the answer.

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

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Answer 5

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

(c)

To determine

The proper way to get the answer and convince the student.

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

(e)

To determine

The way to get the answer.

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Answer 6

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Answer 7

Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4(7+3)$ .

Answer 8

(a)

Expert Solution

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ .

The formula to calculate scalar quantity $A$ is,

$A=a(b+c)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$ , $7$ for $b$ and $3$ for $c$ in the above expression.

$A=4(7+3)=40$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4(7+3)$ is $40$ .

Answer 13

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Answer 14

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700i^+3.00k^=Ai^+Bj^+Ck^$ .

Answer 15

(b)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The given vectors are,

$700i^+3.00k^$ and $Ai^7+Bj^+Ck^$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $i^$ , $j^$ and $k^$ ,

$A=700B=0C=3.00$

Conclusion:

Therefore, the value of $A$ is $7$ , $B$ is $0$ and $C$ is $3$ .

Answer 20

(c)

To determine

The proper way to get the answer and convince the student.

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Answer 21

(c)

To determine

The proper way to get the answer and convince the student.

Answer 22

(c)

Expert Solution

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ .

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $i^$ in $x$ direction $j^$ in $y$ direction and $k^$ in $z$ direction.

The coefficients of $i^$ , $j^$ and $k^$ are compared to get the three unknowns.

The coefficient of $j^$ is $0$ in $700i^+3.00k^$ . So the value of $B$ is $0.$ The coefficient of $i^$ is $700$ in $700i^+3.00k^$ . So the value of $A$ is $700$ and the coefficient of $k^$ is $3.00$ in $700i^+3.00k^$ . So the value of $C$ is $3.00$ .

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Answer 27

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Answer 28

(d)

To determine

The values of $A$ , $B$ , $C$ , $D$ and $E$ from the functional equality.

Answer 29

(d)

Expert Solution

Answer to Problem 16.42P

The value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ , $D$ is $4 rad/s$ and $E$ is $2 rad$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

Compare the given functional equality for the values of $A$ , $B$ , $C$ , $D$ and $E$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Rewrite the above equation as,

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the above expression.

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the value of $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Answer 34

(e)

To determine

The way to get the answer.

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Answer 35

(e)

To determine

The way to get the answer.

Answer 36

(e)

Expert Solution

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given info: The scalar quantity $A$ is $4(7+3)$ and the vector quantity is $700i^+3.00k^=Ai^+Bj^+Ck^$ . The functional equality is $A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$ .

The given functional equality is,

$A+Bcos(Cx+Dt+E)=7.00cos(3x+4t+2)$

Add $0$ on the $R.H.S$ of the above equation.

$A+Bcos(Cx+Dt+E)=0+7.00cos(3x+4t+2)$

Compare the R.H.S and L.H.S of the above equality,

$A=0B=7.00C=3 rad/mD=4 rad/sE=2 rad$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$ , $B$ is $7.00$ , $C$ is $3 rad/m$ and $D$ is $4 rad/s$ .