/*! 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.28. While vacationing in the mountai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Q.28. (page 78)

While vacationing in the mountains you do some hiking. In the morning, your displacement is S鈫�morning=(2000m, east )+(3000m, north )+(200m, vertical \()\). After lunch, your displacement is S鈫�afternoon=(1500m, west )+(2000m, north )鈭�(300m, vertical).

a. At the end of the hike, how much higher or lower are you compared to your starting point?

b. What is the magnitude of your net displacement for the day?

Short Answer

Expert verified

a. The person is100mlower from the starting point.

b. The magnitude of the net displacement of the person is 5.0km.

Step by step solution

01

Part.a.

Use the concept of vector addition to find the net displacement of the person during the hike for the day.

The vertical component of the net displacement vector gives how much the person higher or lower from its starting point and the negative sign indicates that the person is lower from it starting point. Therefore, the person is 100mlower from the starting point.

02

Part.b.

The magnitude of the net displacement is given as follows:

S鈫�net=(500m)2+(5000m)2+(鈭�100m)2=5.0脳103m1km103m=5.0km

Therefore, the magnitude of the net displacement of the person is5.0km.

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 91影视!

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

45. II FIGURE P3.45 shows four electric charges located at the corners of a rectangle. Like charges, you will recall, repel each other while opposite charges attract. Charge B exerts a repulsive force (directly away from B) on charge A of 3.0N. Charge C exerts an attractive force (directly toward C) on charge A of 6.0N. Finally, charge D exerts an attractive force of 2.0Non charge A. Assuming that forces are vectors, what are the magnitude and direction of the net forceF鈫�net exerted on charge A?

Let A鈫�=4i^鈭�2j^, B鈫�=鈭�3i^+5j^and C=A+B.

a. Write vector C鈫�in component form.

b. Draw a coordinate system and on it show vectorsA鈫�,B鈫�, andC鈫�.

c. What are the magnitude and direction of vector C鈫�?

5. I A position vector in the first quadrant has an x-component of 6m and a magnitude of 10m . What is the value of its y-component?

II Ruth sets out to visit her friend Ward, who lives north and east of her. She starts by driving east, but after she comes to a detour that takes her south before going east again. She then drives east for and runs out of gas, so Ward flies there in his small plane to get her. What is Ward's displacement vector? Give your answer (a) in component form, using a coordinate system in which the -axis points north, and (b) as a magnitude and direction.

II Draw each of the following vectors, label an angle that specifies the vector's direction, then find its magnitude and direction.

a.B=鈭�4i^+4j^

b.r鈫�=(鈭�2.0i^鈭�1.0j^)cm

c.v鈫�=(鈭�10i^鈭�100j^)m/s

d.a=(20i^+10j^)m/s2
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Electromagnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Scientific Method Physics

Read Explanation

Electrostatics

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.