/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Vector \(\vec{A}\) has \(y\) -co... [FREE SOLUTION] | 91Ó°ÊÓ

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Vector \(\vec{A}\) has \(y\) -component \(A_{y}=+13.0 \mathrm{m} . \vec{A}\) makes an angle of \(32.0^{\circ}\) counterclockwise from the \(+y\) -axis. (a) What is the \(x\) -component of \(\vec{A} ?\) (b) What is the magnitude of \(\vec{A} ?\)

Short Answer

Expert verified
(a) \(A_x \approx 8.12 \, \text{m}\); (b) Magnitude \(A \approx 15.33 \, \text{m}\).

Step by step solution

01

Understanding the Problem

We are given a vector \(\vec{A}\) with a \(y\)-component, \(A_y = +13.0 \, \text{m}\), and an angle of \(32.0^{\circ}\) from the \(+y\)-axis. We need to determine the \(x\)-component \(A_x\) and the magnitude \(A\) of the vector.
02

Find the x-component (A_x)

The vector \(\vec{A}\) forms a right triangle with its \(x\) and \(y\)-components. The angle between the vector and the \(+y\)-axis is \(32.0^{\circ}\). The \(x\)-component can be found using the tangent function, since we know the \(y\)-component:\[ \tan(\theta) = \frac{A_x}{A_y} \]Substitute \(\theta = 32.0^{\circ}\) and \(A_y = 13.0 \, \text{m}\):\[ \tan(32.0^{\circ}) = \frac{A_x}{13.0} \]\[ A_x = 13.0 \times \tan(32.0^{\circ}) \]Calculating this gives:\[ A_x \approx 13.0 \times 0.6249 \approx 8.12 \, \text{m} \]
03

Find the Magnitude (A)

The magnitude of a vector can be found using Pythagoras' theorem in a right triangle:\[ A = \sqrt{A_x^2 + A_y^2} \]Now plug in the values we have, \(A_x = 8.12 \, \text{m}\) and \(A_y = 13.0 \, \text{m}\):\[ A = \sqrt{(8.12)^2 + (13.0)^2} \]\[ A = \sqrt{65.95 + 169.0} \]\[ A \approx \sqrt{234.95} \approx 15.33 \, \text{m} \]

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Key Concepts

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

Vector Magnitude
The vector magnitude is a measure of how long a vector is. It essentially tells you the "size" or "length" of the vector. To visualize this, consider a vector as an arrow pointing in some direction. The magnitude is like the length of the arrow.

To find the magnitude of a vector, particularly in two dimensions where it has both an x and a y component, you can use the Pythagorean theorem — more on that later. In our context, since the vector components are given in a right triangle formation, the calculation requires squaring both the x and y components, adding them, and then taking the square root of that sum.

In the original problem, the magnitude of vector \( \vec{A} \) is calculated as follows:
  • Calculate \( A_x^2 \) and \( A_y^2 \)
  • Add them: \( A_x^2 + A_y^2 \)
  • Take the square root: \( A = \sqrt{A_x^2 + A_y^2} \)
Using the values \( A_x = 8.12 \, \text{m} \) and \( A_y = 13.0 \, \text{m} \), the magnitude is found to be approximately 15.33 meters.
Right Triangle
A right triangle is a triangle in which one of the angles is exactly 90 degrees. This concept is foundational in trigonometry and vector analysis. Each side of a right triangle plays a specific role: one is the hypotenuse and the others are the opposite and adjacent sides relative to a specified angle.

In the context of vectors, the components (x and y) of a vector can be seen as forming a right triangle, with the vector itself being the hypotenuse. This relationship is key for understanding how vector components relate to their corresponding vector.
  • The x-component and y-component are perpendicular, forming the two short sides of the right triangle.
  • The original vector is like the hypotenuse of this triangle.
Understanding this setup makes it easier to apply trigonometric functions and the Pythagorean theorem to find unknown vector components.
Trigonometric Functions
Trigonometric functions help us relate the angles and sides of a right triangle. In vector problems, they can be used to break down vectors into their components, find missing angles, or determine vector magnitudes.

Key trigonometric functions include:
  • Sine (sin), which relates the length of the opposite side to the hypotenuse.
  • Cosine (cos), which relates the adjacent side to the hypotenuse.
  • Tangent (tan), which relates the opposite side to the adjacent side.
In the original exercise, the tangent function \( \tan(\theta) = \frac{A_x}{A_y} \) was used to find the x-component of the vector given the angle and the y-component. This highlights how trigonometry facilitates converting between angular conflict and linear measurements.
Pythagorean Theorem
The Pythagorean theorem is a cornerstone of geometry and is frequently used in problems involving right triangles. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides.

Mathematically, it is expressed as:
  • For a triangle with sides \( a \) and \( b \), and hypotenuse \( c \), the theorem is \( a^2 + b^2 = c^2 \).
In vector problems, you can use the Pythagorean theorem to find the magnitude of a vector by treating the vector components as the two shorter sides of a right triangle:

\[ c = \sqrt{a^2 + b^2} \]

In the solution provided, this theorem is applied to find the magnitude of the vector \( \vec{A} \) with the given x and y components. It's an elegant way to quantify vector lengths from their components, underscoring the powerful symbiosis between algebra and geometry.

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