To Choose: The correct option.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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

Problem 04.08 (17 points). Answer the following questions related to the figure below. ථි R₁ www R₂ E R₁ www ли R₁ A Use Kirchhoff's laws to calculate the currents through each battery and resistor in terms of R1, R2, E1, & E2. B Given that all the resistances and EMFs have positive values, if E₁ > E2 and R₁ > R2, which direction is the current flowing through E₁? Through R₂? C If E1 E2 and R₁ > R2, which direction is the current flowing through E₁? Through R2?

A 105- and a 45.0-Q resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.268 A. When the 45.0-resistor is disconnected, the current from the battery drops to 0.0840 A. Determine (a) the emf and (b) the internal resistance of the battery. 10 R2 R₁ ww R₁ Emf 14 Emf Final circuit Initial circuit

A ball is shot at an angle of 60° with the ground. What should be the initial velocity of the ball so that it will go inside the ring 8 meters away and 3 meters high. Suppose that you want the ball to be scored exactly at the buzzer, determine the required time to throw and shoot the ball. Full solution and figure if there is.

Answer 2

Question

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

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 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

Answer 6

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

Answer 7

Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Answer 8

Expert Solution & Answer

Answer to Problem 2P

Option (b)

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Two nonzero vectors A→ and B→ .

Formula used:

The formula of vector product:

A→×B→=|A→||B→|sinθ n^

·|A→|= Magnitude of vector B

·|A→|= Magnitude of vector A

·θ= Angle between the vectors A→ and B→ .

·n^= Unit vector perpendicular to the plane A→×B→ .

Calculation:

When A→ and B→ are parallel:

θ=0o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin0o n^=|A→||B→|×0 n^ ∵(sin0o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are perpendicular:

θ=90o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin90o n^=|A→||B→|×1 n^ ∵(sin 90o=1)=|A→||B→| n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | )2=|A→||B→|

When A→ and B→ are antiparallel:

θ=180o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin180o n^=|A→||B→|×0 n^ ∵(sin 180o=0)=0 n^

The magnitude of the vector product:

|A→×B→|=02=0

When A→ and B→ are at 45o :

θ=45o

A→×B→=|A→||B→|sinθ n^=|A→||B→|sin45o n^=|A→||B→|×12 n^ ∵(sin 90o=1)=| A →|| B →|2 n^

The magnitude of the vector product:

|A→×B→|= ( coefficient of vector )2= ( | A → || B → | 2 )2=| A →|| B →|2

The greatest magnitude of the vector product is |A→||B→| .

Conclusion:

Hence, when the vectors A→ and B→ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.

Answer 12

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

To Choose: The correct option.

To Choose: The correct option.

Answer to Problem 2P

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions