/*! 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 1 What are (a) the \(x\) component... [FREE SOLUTION] | 91Ó°ÊÓ

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What are (a) the \(x\) component and (b) the \(y\) component of a vector \(\vec{a}\) in the \(x y\) plane if its direction is \(250^{\circ}\) counterclockwise from the positive direction of the \(x\) axis and its magnitude is \(7.3 \mathrm{~m} ?\)

Short Answer

Expert verified
The x component is approximately -2.5 m, and the y component is approximately -6.9 m.

Step by step solution

01

Understand the Direction of the Vector

The vector \(\vec{a}\) has a direction given as \(250^{\circ}\) counterclockwise from the positive \(x\) axis. This direction implies that the vector is in the third quadrant of the Cartesian coordinate system, since \(250^{\circ}\) falls between \(180^{\circ}\) and \(270^{\circ}\).
02

Calculate the Angles with Axes

To find the components of the vector, we need to determine the angle \(\theta\) it forms with the negative \(x\) axis. Since the negative \(x\) axis is at \(180^{\circ}\), the angle \(\theta'\) with the positive \(x\) axis is \(250^{\circ} - 180^{\circ} = 70^{\circ}\).
03

Calculate the X Component of the Vector

The \(x\) component of a vector \(\vec{a}\) is given by \(a_x = a \cdot \cos(\theta')\). Given \(a = 7.3\) m and \(\theta' = 250^{\circ}\), we have: \[ a_x = 7.3 \cdot \cos(250^{\circ}) = 7.3 \cdot \cos(70^{\circ}) \approx 7.3 \cdot 0.3420 \approx 2.5\] Note that the actual \(x\) component will be negative due to the third quadrant location: \(a_x \approx -2.5\).
04

Calculate the Y Component of the Vector

The \(y\) component of a vector \(\vec{a}\) is given by \(a_y = a \cdot \sin(\theta')\). Using \(a = 7.3\) m and \(\theta' = 250^{\circ}\), we have: \[ a_y = 7.3 \cdot \sin(250^{\circ}) = 7.3 \cdot \sin(70^{\circ}) \approx 7.3 \cdot 0.9397 \approx 6.9\] Note that the actual \(y\) component will also be negative due to the third quadrant location: \(a_y \approx -6.9\).

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Key Concepts

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

Cartesian Coordinate System
The Cartesian Coordinate System is a two-dimensional plane used to pinpoint locations through two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). This system allows us to define vectors in a straightforward manner using coordinates. When dealing with vectors, such as vector \( \vec{a} \), we often describe them in terms of their components along these axes. Each component is a projection of the vector onto one of the axes.

In this exercise, the vector is placed in the \(xy\)-plane, meaning its full description requires both an \(x\) and a \(y\) component. The concept of quadrants is central; these are the four distinct sections of the plane determined by the axes' intersection. Here, with the vector at an angle of \(250^{\circ}\), it exists in the third quadrant. This implies that both components of the vector will be negative since both the \(x\) and the \(y\) directions point towards negative infinity in this quadrant.
  • Positive x-axis = 0°
  • Positive y-axis = 90°
  • Negative x-axis = 180°
  • Negative y-axis = 270°
Understanding which quadrant a vector lands in is crucial for correctly determining the signs of its components.
Trigonometric Functions
Trigonometric functions, notably sine (\(\sin\)) and cosine (\(\cos\)), are pivotal in determining the components of a vector based on its direction and magnitude.When we describe a vector's angle from the positive x-axis, we aim to untangle how much of this vector points horizontally and vertically. This is where trigonometry steps in.
  • The x-component of a vector is found by multiplying the magnitude of the vector by \( \cos(\theta') \), where \( \theta' \) is its angle with respect to the horizontal direction (after subtracting any full circle rotations if needed).
  • The y-component is calculated using the magnitude multiplied by \( \sin(\theta') \).
For a vector at 250°, the computations utilize \(70°\) (the simplified angle made with the negative x-axis). Knowing these operations and angles help us determine that in this particular case:
  • \(a_x = 7.3 \cdot \cos(70°) \approx -2.5\)
  • \(a_y = 7.3 \cdot \sin(70°) \approx -6.9\)
Trigonometry allows efficient and accurate computations, opening a window to coordinate various mechanics in physics and engineering.
Vector Magnitude and Direction
Every vector \( \vec{a} \) has a magnitude, indicating its length or size, and a direction, which shows where it points. These two properties are interconnected in defining the vector's position in space.

  • The magnitude is denoted as \( |\vec{a}| \) or simply \( a \), representing how long the vector is. In the exercise, it is given as \(7.3 \text{ m}\).
  • The direction of a vector is tied to its angle \( \theta \) relative to a defined prime direction, here the positive x-axis, with counterclockwise movement being positive.
The angle given, \(250°\), illustrates the direction relative to the positive x-axis. This direction can be visualized geometrically or numerically computed using reverse trigonometric functions when provided with components.The combination of magnitude and direction simplifies plotting and integrating vectors into larger structures or systems, such as forces acting on an object or an object’s velocity in space. By breaking down vectors into their axial components, using magnitude and direction, navigable solutions and analyses become clear and accessible, suited for both academic study and practical applications.

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Most popular questions from this chapter

\(\vec{A}\) has the magnitude \(12.0 \mathrm{~m}\) and is angled \(60.0^{\circ}\) counterclockwise from the positive direction of the \(x\) axis of an \(x y\) coordinate system. Also, \(\vec{B}=(12.0 \mathrm{~m}) \hat{\mathrm{i}}+(8.00 \mathrm{~m}) \hat{\mathrm{j}}\) on that same coordinate system. We now rotate the system counterclockwise about the origin by \(20.0^{\circ}\) to form an \(x^{\prime} y^{\prime}\) system. On this new system, what are (a) \(\vec{A}\) and (b) \(\vec{B}\), both in unit-vector notation?

(a) What is the sum of the following four vectors in unitvector notation? For that sum, what are (b) the magnitude, (c) the angle in degrees, and (d) the angle in radians? \(\vec{E}: 6.00 \mathrm{~m}\) at \(+0.900 \mathrm{rad} \quad \vec{F}: 5.00 \mathrm{~m}\) at \(-75.0^{\circ}\) \(\vec{G}: 4.00 \mathrm{~m} \mathrm{at}+1.20 \mathrm{rad}\) \(\vec{H}: 6.00 \mathrm{~m}\) at \(-21.0^{\circ}\)

Three vectors \(\vec{a}, \vec{b},\) and \(\vec{c}\) each have a magnitude of \(50 \mathrm{~m}\) and lie in an \(x y\) plane. Their directions relative to the positive direction of the \(x\) axis are \(30^{\circ}, 195^{\circ},\) and \(315^{\circ},\) respectively. What are (a) the magnitude and (b) the angle of the vector \(\vec{a}+\vec{b}+\vec{c}\), and (c) the magnitude and (d) the angle of \(\vec{a}-\vec{b} \pm \vec{c}\) ? What are the (e) magnitude and (f) angle of a fourth vector \(\vec{d}\) such that \((\vec{a}+\vec{b})-(\vec{c}+\vec{d})=0 ?\)

Express the following angles in radians: (a) \(20.0^{\circ},\) (b) \(50.0^{\circ}\), (c) \(100^{\circ} .\) Convert the following angles to degrees: (d) \(0.330 \mathrm{rad}\), (e) \(2.10 \mathrm{rad},\) (f) 7.70 rad.

For the following three vectors, what is \(3 \vec{C} \cdot(2 \vec{A} \times \vec{B}) ?\) \(\vec{A}=2.00 \hat{\mathrm{i}}+3.00 \hat{\mathrm{j}}-4.00 \hat{\mathrm{k}}\) \(\vec{B}=-3.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}} \quad \vec{C}=7.00 \hat{\mathrm{i}}-8.00 \hat{\mathrm{j}}\)

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