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The scalar product of vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is \(+\)48.0 m\(^2\). Vector \(\overrightarrow{A}\) has magnitude 9.00 m and direction 28.0\(^{\circ}\) west of south. If vector \(\overrightarrow{B}\) has direction 39.0\(^{\circ}\) south of east, what is the magnitude of \(\overrightarrow{B}\)?

Short Answer

Expert verified
Vector \(\overrightarrow{B}\) has a magnitude of approximately 27.9 meters.

Step by step solution

01

Understanding the Dot Product

The dot product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) is given by \( \overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos{\theta} \), where \(\theta\) is the angle between the two vectors. From the problem, we know this dot product equals 48 m\(^2\).
02

Calculating the Angle Between the Vectors

The angle given in the problem are with respect to cardinal directions. Vector \( \overrightarrow{A} \) is 28.0\(^{\circ}\) west of south, which translates to 28.0\(^{\circ}\) + 90.0\(^{\circ}\) = 118.0\(^{\circ}\) from the east direction. Vector \( \overrightarrow{B} \) is 39.0\(^{\circ}\) south of east. The angle \(\theta\) between the vectors is \(118.0^{\circ} - 39.0^{\circ} = 79.0^{\circ}\).
03

Apply Dot Product Formula

Substituting the known values into the dot product formula, we have:\[ 48.0 = 9.00 \times |\overrightarrow{B}| \times \cos(79.0^{\circ}) \]
04

Solving for Magnitude of Vector B

First, calculate \(\cos(79.0^{\circ})\) using a calculator to find it approximately equal to 0.1908. Then solve the equation:\[ 48.0 = 9.00 \times |\overrightarrow{B}| \times 0.1908 \]\[ |\overrightarrow{B}| = \frac{48.0}{9.00 \times 0.1908} \]\[ |\overrightarrow{B}| \approx 27.9 \text{ meters} \]
05

Verification

Verify your calculations to ensure accuracy. You can double-check the computations of \(\cos(79.0^{\circ})\) and the division to confirm that \(|\overrightarrow{B}| \approx 27.9 \text{ meters}\) is consistent throughout the steps.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnitude of a Vector
The magnitude of a vector can be thought of as the length or size of the vector. It's like measuring a straight-line distance from one point to another. This is important because, in physics, vectors have both a magnitude and a direction.

When you need to find the magnitude of a vector, you often use the formula \(|\overrightarrow{V}| = \sqrt{V_x^2 + V_y^2}\). This applies to vectors in a two-dimensional space where \(V_x\) and \(V_y\) are the vector's components along the x and y axes. However, if you know the direction and other relationships between vectors (as in the dot product), you may not need this formula directly.

In our example, you calculated the magnitude of vector \(\overrightarrow{B}\) using the scalar (dot) product and angles involved. It shows how a vector's magnitude can be derived from its relationship with other vectors rather than its components.
Angle Between Vectors
The angle between vectors is key to understanding their relationship. It tells us how the vectors are oriented towards each other. In trigonometry, this angle is fundamental because it helps calculate many vector operations, like the dot product.

Here, you express the angle using the directions provided in the problem. Vectors can be described with angles relative to cardinal directions (north, south, east, west), and converting these can clarify the angle between the vectors. In this exercise, you found the relevant angle using simple subtraction of these converted angles to reach the value of \(79.0^{\circ}\).

Understanding these angles becomes crucial when applying trigonometric functions, as they determine functions like \(\cos(\theta)\), extensively used in vector mathematics to find products or projections.
Trigonometry in Physics
Trigonometry connects deeply with physics because it provides tools to manipulate and understand vectors, which represent forces, velocities, and displacements in the physical world.
  • Through trigonometry, you can decompose vectors into components to better analyze them.
  • Using functions like cosine and sine, you quantify relationships between vector magnitudes and angles.
In the given problem, trigonometry is pivotal for cracking the dot product equation. The cosine of the angle (\(\cos(79.0^{\circ})\)) is essential because it's the trigonometric function that relates the scalar product to the magnitudes of the vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\).

Utilizing trigonometric identities and functions allows us to handle more complex interactions between vectors, predicting outcomes in fields like engineering or physics efficiently. By understanding these concepts, physics problems become simpler and more intuitive to resolve.

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Most popular questions from this chapter

While following a treasure map, you start at an old oak tree. You first walk 825 m directly south, then turn and walk 1.25 km at 30.0\(^{\circ}\) west of north, and finally walk 1.00 km at 32.0\(^{\circ}\) north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reasonable, compare it with a graphical solution drawn roughly to scale.

In the methane molecule, CH\(_4\), each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the center. In coordinates for which one of the C\(-\)H bonds is in the direction of \(\hat{\imath}\) + \(\hat{\jmath}\) + \(\hat{k}\), an adjacent C\(-\)H bond is in the \(\hat{\imath}\) \(-\) \(\hat{\jmath}\) \(-\) \(\hat{k}\) direction. Calculate the angle between these two bonds.

Later in our study of physics we will encounter quantities represented by (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\). (a) Prove that for any three vectors \(\overrightarrow{A}\), \(\overrightarrow{B}\), and \(\overrightarrow{C}\), \(\overrightarrow{A}\) \(\cdot\) (\(\overrightarrow{B}\) \\(\times\\) \(\overrightarrow{C}\)) = (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\). (b) Calculate (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\) for vector \(\overrightarrow{A}\) with magnitude \(A\) = 5.00 and angle \(\theta$$_A\) = 26.0\(^{\circ}\) (measured from the +\(x\)-axis toward the \(+$$y\)-axis), vector \(\overrightarrow{B}\) with \(B\) = 4.00 and \(\theta$$_B\) = 63.0\(^{\circ}\), and vector \(\overrightarrow{C}\) with magnitude 6.00 and in the \(+$$z\)-direction. Vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are in the \(x$$y\)-plane.

As a test of orienteering skills, your physics class holds a contest in a large, open field. Each contestant is told to travel 20.8 m due north from the starting point, then 38.0 m due east, and finally 18.0 m in the direction 33.0\(^{\circ}\) west of south. After the specified displacements, a contestant will find a silver dollar hidden under a rock. The winner is the person who takes the shortest time to reach the location of the silver dollar. Remembering what you learned in class, you run on a straight line from the starting point to the hidden coin. How far and in what direction do you run?

Find the magnitude and direction of the vector represented by the following pairs of components: (a) A\(x\) = \(-\)8.60 cm, A\(y\) = 5.20 cm; (b) A\(x\) = \(-\)9.70 m, A\(y\) = -2.45 m; (c) A\(x\) = 7.75 km, A\(y\) = \(-\)2.70 km.

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