/*! 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} Q. 24 a. What is the angle ϕ between... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

a. What is the angle ϕbetween vectors E→andF→in FIGURE P3.24?

b. Use geometry and trigonometry to determine the magnitude and direction of G→=E→+F→.

c. Use components to determine the magnitude and direction of G→=E→+F→.

Short Answer

Expert verified

(a) ϕ=71.6∘.

(b) The magnitude is 3and the direction is 45∘.

(c) The magnitude is3and the direction is90∘.

Step by step solution

01

Part (a) Step 1. Given information

The given figure is,

We need to determine the magnitude and direction ofG→=E→+F→by using geometry and trigometry.

02

Step 2. Calculation

The magnitude and the direction of G→can be determined by the cosine rule.

G→2=2+5-225cos180-ϕ.

=2+5-215cos180-71.6.

≈9⇒G→≈3.

The direction can be determined by using an angle between Gand E.

cosα=32+2-5232.

cosα=12.

α=cos-112.

=45∘.=45∘.

03

Part (c) Step 1. Given information

The given figure is,

We need to determine the magnitude and direction ofG→=E→+F→by using the components.

04

Step 2. Calculation

G→=E→+F→.

role="math" =1+-1i^+1+2j^.

=3j^.

G→=02+32.

=9.

role="math" localid="1649272859822" Ï•=3.

tanθ=GyGx.

role="math" =30.

θ=tan-10.

=90∘.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.