/*! 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 30 Find the magnitude and direction... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The magnitude and direction of the vectors are: (a) 10 cm and 149°, (b) 10 m and 194°, (c) 8 km and 341°.

Step by step solution

01

Find the magnitude of the vectors

The magnitude of a vector (A) can be found by using the formula \(A = \sqrt{A_x^2 + A_y^2}\) where \(A_x\) and \(A_y\) are the x and y components of the vector respectively. (a) \(A = \sqrt{(-8.60 cm)^2 + (5.20 cm)^2} = 10 cm\)(b) \(A = \sqrt{(-9.70 m)^2 + (-2.45 m)^2} = 10 m\)(c) \(A = \sqrt{(7.75 km)^2 + (-2.70 km)^2} = 8 km\)
02

Find the direction of the vectors

The direction of a vector can be found using the formula \(θ = arctan(A_y/A_x)\), where \(θ\) is the angle that the vector makes with the positive x-axis (measured counterclockwise), and \(A_x\) and \(A_y\) are the x and y components of the vector respectively. Before using this formula, you have to adjust for the correct quadrant.(a) Since \(A_x < 0\) and \(A_y > 0\), \(θ = 180° + arctan(5.20 cm / -8.60 cm) = 149°\)(b) Since both \(A_x < 0\) and \(A_y < 0\), \(θ = 180° + arctan(-2.45 m / -9.70 m) = 194°\)(c) Since \(A_x > 0\) and \(A_y < 0\), \(θ = 360° + arctan(-2.70 km / 7.75 km) = 341°\)

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Key Concepts

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

Understanding Vector Components
Every vector in a two-dimensional plane can be described by two numerical values called its components. These components are usually along the x-axis (horizontal) and y-axis (vertical).
A vector's components tell us how much of the vector lies in each direction. Together, these components can give us a comprehensive picture of the vector's overall direction and magnitude.
Imagine a vector as an arrow pointing from one point to another. The x-component is how much the arrow goes along the horizontal, and the y-component is how much it rises or falls vertically.
  • The x-component is often denoted as \(A_x\).
  • The y-component is often denoted as \(A_y\).
To find the components in practical terms, we might write a vector as \((A_x, A_y)\), which means that the vector has a total reach of \(A_x\) units in the horizontal direction and \(A_y\) units in the vertical. These values can help us determine other properties of the vector, such as its magnitude and direction. As the components suggest, the direction and the length of the arrow are dependent on both components working together.
Calculating Vector Magnitude
The magnitude of a vector tells us how long the vector is, or, how far it extends from its initial point. Think of it as measuring the length of the arrow described by the vector.
Mathematically, the magnitude \(A\) of a vector with components \(A_x\) and \(A_y\) can be found using the Pythagorean theorem:
  • The formula is \(A = \sqrt{A_x^2 + A_y^2}\).
This formula comes from the relationship found in right triangles, where the magnitude is the hypotenuse. Each component \(A_x\) and \(A_y\) represents the legs of a right triangle.
Here’s how it works:1. Square each component \(A_x\) and \(A_y\).2. Add these squares together.3. Take the square root of the result.For example, if a vector has components \(-8.60 \ cm\) and \(5.20 \ cm\), the magnitude is calculated as follows:\[\text{Magnitude} = \sqrt{(-8.60)^2 + (5.20)^2} = 10 \ cm\]Use this method to find how long each vector really is, no matter which directions its components point.
Determining Vector Direction
Vector direction is crucial as it tells us where the vector is pointed. To determine a vector's direction, we often measure the angle it forms with the positive x-axis.
The direction \(\theta\) can be calculated using the arctangent function, specifically \(\theta = \arctan\left( \frac{A_y}{A_x} \right)\). But, there's an important detail: this angle depends on the sign and magnitude of the components, which means the real direction also depends on the vector's quadrant.
  • For vectors in the second quadrant, which have \(A_x < 0\) and \(A_y > 0\), the formula becomes \(\theta = 180^\circ + \arctan\left(\frac{A_y}{A_x}\right)\).
  • For the third quadrant, where \(A_x < 0\) and \(A_y < 0\), use \(\theta = 180^\circ + \arctan\left(\frac{A_y}{A_x}\right)\).
  • For the fourth quadrant, with \(A_x > 0\) and \(A_y < 0\), use \(\theta = 360^\circ + \arctan\left(\frac{A_y}{A_x}\right)\).
It's like finding which way the wind is blowing—you need to know in which direction the power is strongest.

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Most popular questions from this chapter

(a) The recommended daily allowance (RDA) of the trace metal magnesium is \(410 \mathrm{mg} /\) day for males. Express this quantity in \(\mu \mathrm{g} /\) day \(.\) (b) For adults, the RDA of the amino acid lysine is \(12 \mathrm{mg}\) per \(\mathrm{kg}\) of body weight. How many grams per day should a \(75 \mathrm{~kg}\) adult receive? (c) A typical multivitamin tablet can contain \(2.0 \mathrm{mg}\) of vitamin \(\mathrm{B}_{2}\) (riboflavin), and the RDA is \(0.0030 \mathrm{~g} /\) day. How many such tablets should a person take each day to get the proper amount of this vitamin, if he gets none from other sources? (d) The RDA for the trace element selenium is \(0.000070 \mathrm{~g} /\) day. Express this dose in \(\mathrm{mg} /\) day.

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As noted in Exercise 1.23 , a spelunker is surveying a cave. She follows a passage \(180 \mathrm{~m}\) straight west, then \(210 \mathrm{~m}\) in a direction \(45^{\circ}\) east of south, and then \(280 \mathrm{~m}\) at \(30^{\circ}\) east of north. After a fourth displacement, she finds herself back where she started. Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.

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