To Choose: The correct option.

Question

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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.

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SARET CRKS AUTOWAY 12. A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.55 s. How high is the cliff? 13. A ball is dropped from rest at the top of a building that is 320 m tall. Assuming no air resistance, what is the speed of the ball just before it strikes the ground? 14. Estimate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (280m high), and (b) his velocity just before "landing". Useful equations For Constant Velocity: V => D X = V₁t + Xo For Constant Acceleration: Vr = V + at X = Xo+Vot + v=V+2a(X-Xo) \prom = V +V V velocity t = time D Distance X = Final Position Xo Initial Position V = Final Velocity Vo Initial Velocity a = acceleration For free fall Yf = Final Position Yo Initial Position g = 9.80 m $2 For free fall: V = V + gt Y=Yo+Vo t + +gt V,² = V₁²+2g (Y-Yo) V+Vo Vprom= 2 6

Solve the problems

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.

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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.

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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.

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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

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Chapter 10 Solutions