# Sir Eman's Online Physics Quiz No. 02

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| Written by Emankulit
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Emankulit
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Quizzes Created: 3 | Total Attempts: 790
Questions: 15 | Attempts: 463

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Sir Eman's Online Physics Quiz No. 02

Topic: Vectors and Vector Multiplication
Time Limit: 30 minutes

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• 1.

### The magnitude of two vectors A and B are A = 12 units and B = 8 units. Which of the following pairs of numbers represents the largest and smallest possible values for the magnitude of the resultant vector R = A + B?

• A.

14.4 units , 4 units

• B.

12 units, 8 units

• C.

20 units, 4 units

• D.

None of these answers

C. 20 units, 4 units
Explanation
ang obvious naman kung bakit yun ang sagot di'ba? nyahahaha!

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• 2.

### Given the vectors A = (3i - 4j + 4k) and B = (2i + 3j - 7k), find the magnitude of the vector C = A + B. (3 significant figures)

5.92
5.93
5.91
5.90
5.94
Explanation
The magnitude of a vector is calculated by taking the square root of the sum of the squares of its components. In this case, the vector C = A + B is obtained by adding the corresponding components of vectors A and B.

C = (3i - 4j + 4k) + (2i + 3j - 7k)
= (3 + 2)i + (-4 + 3)j + (4 - 7)k
= 5i - j - 3k

To find the magnitude of C, we calculate:
|C| = √((5)^2 + (-1)^2 + (-3)^2)
= √(25 + 1 + 9)
= √35

Rounding to three significant figures, the magnitude of C is approximately 5.92.

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• 3.

### Given the vectors A = (3i - 4j + 4k) and B = (2i + 3j - 7k), find the magnitude of the vector C = 2A - B. (3 significant figures)

19.0
Explanation
To find the magnitude of vector C, we first need to calculate the vector C by subtracting vector B from twice vector A. Multiplying vector A by 2 gives us (6i - 8j + 8k), and subtracting vector B gives us (6i - 8j + 8k) - (2i + 3j - 7k) = 4i - 11j + 15k. The magnitude of vector C can be found using the formula sqrt((4^2) + (-11^2) + (15^2)), which simplifies to sqrt(16 + 121 + 225) = sqrt(362). Rounding to three significant figures gives us 19.0.

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• 4.

### It is a scalar that is the product of the magnitude of A multiplied to the component of B along A.

Dot Product
Scalar Product
Explanation
The dot product, also known as the scalar product, is a mathematical operation that involves multiplying the magnitudes of two vectors and the cosine of the angle between them. In this case, the correct answer is "Dot Product, Scalar Product" because the given statement describes the dot product precisely. It is a scalar quantity obtained by multiplying the magnitude of vector A with the component of vector B along vector A.

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• 5.

### It is another vector whose magnitude is equal to the area of the parallelogram formed by A and B and directed perpendicular to both A and B in the Right-hand method.

Vector Product
Cross Product
Explanation
The vector product, also known as the cross product, refers to the mathematical operation between two vectors that results in a vector perpendicular to both input vectors. In this case, the explanation is stating that the correct answer is the vector product or cross product because it produces another vector with a magnitude equal to the area of the parallelogram formed by vectors A and B. This new vector is also directed perpendicular to both A and B according to the Right-hand method.

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• 6.

### Which is the correct way of denoting the Vector product of two vectors A and B?

• A.

A · B

• B.

A (B)

• C.

A x B

• D.

A * B

C. A x B
Explanation
The correct way of denoting the Vector product of two vectors A and B is A x B.

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• 7.

### Which is the correct way of denoting the Scalar product of two vectors A and B?

• A.

A * B

• B.

A · B

• C.

A x B

• D.

A (B)

B. A · B
Explanation
The correct way of denoting the scalar product of two vectors A and B is A · B.

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• 8.

### For the two vectors A = 3i + 2j and B = 4i + 5k. Find the scalar product A · B. (3 significant figures)

12.0
12
12.1
Explanation
The scalar product, also known as the dot product, of two vectors is calculated by multiplying their corresponding components and then summing them up. In this case, the scalar product of vectors A and B can be found by multiplying the corresponding components of A and B, which are 3 and 4 respectively, and then summing them up. This gives us 3 * 4 = 12. Therefore, the scalar product of A and B is 12.

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• 9.

### For the two vectors A = 3i + 2j and B = 4i + 5k. Find the Vector product A x B.

10i - 15j - 8k
10i-15j-8k
10i -15j -8k
Explanation
The vector product of two vectors A and B is calculated using the cross product formula. In this case, A = 3i + 2j and B = 4i + 5k. To find the vector product A x B, we can use the formula:

A x B = (AyBz - AzBy)i - (AxBz - AzBx)j + (AxBy - AyBx)k

Substituting the values, we get:

A x B = (2 * 5 - 0 * 0)i - (3 * 0 - 0 * 4)j + (3 * 0 - 2 * 4)k
= 10i - 0j - 8k
= 10i - 8k

Therefore, the correct answer is 10i - 8k.

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• 10.

### If the component vector A along the direction of vector B is zero, what can you conclude about the two vectors?

• A.

Vectors A and B are parallel to each other.

• B.

Vectors A and B are anti-parallel to each other.

• C.

Vectors A and B are perpendicular to each other.

• D.

Vectors A and B are at an angle with respect to the other.

C. Vectors A and B are perpendicular to each other.
Explanation
If the component vector A along the direction of vector B is zero, it means that A and B are perpendicular to each other. This is because the component of A along B is determined by projecting A onto the direction of B. If this component is zero, it implies that A does not have any projection along B, indicating that A and B are at a right angle or perpendicular to each other.

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• 11.

5, 7
5,7
• 12.

### Find the angle between the following two vectors:A = -2.00i + 6.00jB = 2.00i - 3.00j(three significant figures)

165
Explanation
The angle between two vectors can be found using the dot product formula: A · B = |A| |B| cos(θ), where A and B are the magnitudes of the vectors A and B, and θ is the angle between them. By substituting the given values, we get (-2)(2) + (6)(-3) = sqrt((-2)^2 + 6^2) sqrt((2)^2 + (-3)^2) cos(θ). Simplifying this equation gives us -4 - 18 = sqrt(40) sqrt(13) cos(θ). Further simplification leads to -22 = sqrt(520) cos(θ). Dividing both sides by sqrt(520) gives us cos(θ) = -22/sqrt(520). Taking the inverse cosine of both sides gives us θ = 165 degrees.

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• 13.

### Find the angle between the following two vectors: A = 3.00i + 5.00j B = 10.00i + 6.00j(three significant figures)

28.0
28.07
28
28.1
Explanation
The angle between two vectors can be found using the dot product formula. The dot product of two vectors A and B is equal to the magnitude of A multiplied by the magnitude of B, multiplied by the cosine of the angle between them. By calculating the dot product of A and B, and dividing it by the product of their magnitudes, we can find the cosine of the angle. Taking the inverse cosine of this value will give us the angle between the vectors. In this case, the angle between vectors A and B is approximately 28.0 degrees, as well as 28.07, 28, and 28.1 degrees.

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• 14.

### Find the angle between the following two vectors: A = -4.00i + 2.00j B = 7.00i + 14.00j (three significant figures)

90.0
90
Explanation
The angle between two vectors can be found using the dot product formula: cos(theta) = (A dot B) / (|A| * |B|). In this case, A dot B = (-4)(7) + (2)(14) = 0, |A| = sqrt((-4)^2 + 2^2) = sqrt(20) = 4.47, and |B| = sqrt(7^2 + 14^2) = sqrt(245) = 15.65. Plugging these values into the formula, we get cos(theta) = 0 / (4.47 * 15.65) = 0. Therefore, the angle between the two vectors is 90 degrees.

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• 15.

### Given  two vectors A = 4.00i + 3.00j and B = 5.00i - 2.00j, find the magnitude and direction of A x B . note: express answer in terms of unit vectors and must be in 3 significant figures. must be simplified already

-23.0k
-23k
-23.0 k
-23 k
Explanation
The magnitude and direction of the cross product A x B can be found using the formula |A x B| = |A| |B| sinθ, where |A| and |B| are the magnitudes of vectors A and B, and θ is the angle between them. In this case, |A| = √(4.00^2 + 3.00^2) = 5.00 and |B| = √(5.00^2 + (-2.00)^2) = 5.39. The angle between A and B can be found using the dot product: A · B = |A| |B| cosθ. Solving for θ, we get θ = arccos((A · B) / (|A| |B|)) = arccos((4.00 * 5.00 + 3.00 * (-2.00)) / (5.00 * 5.39)) ≈ 24.1°. Therefore, the magnitude and direction of A x B is -23.0k, which means the vector points in the negative k direction with a magnitude of 23.0.

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