/*! 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} Q18P (a) In figure 1.58, what are the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) In figure 1.58, what are the components of the vector d→?

Short Answer

Expert verified

(a) The component of the vector d→=-6,3,2.

Step by step solution

01

Vector components

The vector has three components. These three components are the unit cut by the vector on x, y and z axis. The vector components can be represented by the following manner.

Let the vector is r, then the vector component of the vector r will be,

r→=rx,ry,rz

Where, rxisxaxisunit,ryisyunit,rziszcomponentunit.

02

(a) Component of vector d→

All the vectors have zero z component here because the z dimension is absent. The component for the vector a covering the 4 unit of x axis and 3 unit of y axis in 3rd coordination centre of the graph hence the component of the vector d→=-6,3,2.

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

Most popular questions from this chapter

In order to pull a sled across a level field at a constant velocity you have to exert a constant force. Doesn’t this violate Newton’s first and second laws of motion, which imply that no force is required to maintain a constant velocity? Explain this seeming contradiction.

At a certain instant a particle is moving in the +xdirection with momentum+8kg·m/s. During the next 0.13sa constant force acts on the particle, with Fx=-7N and xFy=+5N. What is the magnitude of the momentum of the particle at the end of this v interval?

Question: An electron passes location (0.02,0.04,-0.06)m, and 2μslater is detected at location (0.02,1.84,-0.86)m, (1 microsecond is1×10-6s). a) What is the average velocity of the electron? b) If the electron continues to travel at this average velocity, where will it be in another 5μs?

Question: the Hall effect can be used to determine the sign of the mobile charges in a particular conducting material. A bar of a new kind of conducting material is connected to a battery as shown in Figure 20.85. In this diagram, the x-axis runs to the right, the y-axis runs up, and the z-axis runs out of the page, toward you. A voltmeter is connected across the bar as shown, with the leads placed directly opposite each other along a vertical line. In order to answer the following question, you should draw a careful diagram of the situation, including all relevant charges, electric fields, magnetic fields, and velocities.

Initially, there is no magnitude filed in the region of the bar. (a) Inside the bar, what is the direction of the electric field E→due to the charges on the batteries and the surface of the wires and the bar? This is the electric field that drives the current in the bar. (b) If the mobile charges in the bar are positive in what direction do they move when the current runs? (c) If the mobile charges in the bar are negative, in what direction do they move when the current runs? (d) In this situation (zero magnetic fields), what is the sign of the reading on the voltmeter?

Next, large coils (not shown) are moved near the bar. And current runs through the coils, making a magnetic field in the -z direction (into the page). (e) If the mobile charges in the bar are negative, what is the direction of the magnetic force on the mobile charge? (f) If the mobile charges in the bar are negative, which of the following things will happen? (1) Positive charge will accumulate on the top of the bar. (2) The bar will not becomes polarized. (3) Negative charge will accumulate on the left end of the bar. (4) Negative charge will accumulate on the top of the bar. (g) If the mobile charges in the bar are positive, what is the direction of the magnetic force on the mobile charges? (h) If the mobile charges in the bar are positive, which of these things will happen? (1) positive charge will accumulate on the top of the bar. (2) The bar will not becomes polarized. (3) Positive charge will accumulate on the right end of the bar. (4) Negative charge will accumulate on the top of the bar.

You look at the voltmeter and find that the reading on the meter is -5×10-4volts. (i) What can you conclude from this observation? (Remember that a voltmeter gives a positive reading if the positive lead is attached to the higher potential location.) (1) There is not enough information to figure out the sign of the mobile charges. (2) The mobile charges are negative. (3) The mobile charges are positive.

If you did not already to problem P60, do it now. Also calculate numerically the angle through which the yo-yo turns, in radians and degrees.

See all solutions

Recommended explanations on Physics Textbooks

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.