Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

When two point charges of equal mass and charge arereleased on a frictionless table, each has an initial acceleration(magnitude) a0. If instead you keep one fixed and release the otherone, what will be its initial acceleration: a0,2a0,or a0/2? Explain.

Short Answer

Expert verified

The initial acceleration isa0.

Step by step solution

01

Definition of the acceleration

The rate at which an object's velocity changes with function of time is called acceleration.

The acceleration when the velocity is zero is called initial acceleration.

02

Explanation of how the initial acceleration produced

Recognizing Newton's second law explains why it initially speed with a0in the first scenario.

One spot charge accelerates due to a net force exerted on it. The other spot charge is accelerating due to a net force exerted on it. Because the weights and initial maximum acceleration are equal, the net forces are also equal.

In spite of, there is no frictional, the net force is Coulomb's force, which is F=k|q1q2|r2Irepellent because the energies are equal. This initial force applies in opposing directions on both point charges, providing them initial acceleration.

03

Explanation for finding the initial acceleration

See the instant of discharging either both or merely one of the charged objects, and then emphasize the phrase "initial," because this force varies (reduces with distances, as does acceleration).

The force exerted on the charged object is the same as if the other spot charge is liberated or not, therefore it accelerates (at first) at the same pace.

Hence, the initial acceleration is a0.

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 Vaia!

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

(See Discussion Question Q25.14.) An ideal ammeter A is placed in a circuit with a battery and a light bulb as shown in Fig. Q25.15a, and the ammeter reading is noted. The circuit is then reconnected as in Fig. Q25.15b, so that the positions of the ammeter and light bulb are reversed. (a) How does the ammeter reading in the situation shown in Fig. Q25.15a compare to the reading in the situation shown in Fig. Q25.15b? Explain your reasoning. (b) In which situation does the light bulb glow more brightly? Explain your reasoning.

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ωand 800Ω. If the two light bulbs are connected in series across a 120Vline, find (a) the current through each bulb; (b) the power dissipated in each bulb; (c) the total power dissipated in both bulbs. The two light bulbs are now connected in parallel across the120Vline. Find (d) the current through each bulb; (e) the power dissipated in each bulb; (f) the total power dissipated in both bulbs. (g) In each situation, which of the two bulbs glows the brightest? (h) In which situation is there a greater total light output from both bulbs combined?

Batteries are always labeled with their emf; for instances an AA flashlight battery is labelled “ 1.5 V ”. Would it also be appropriate to put a label on batteries starting how much current they provide? Why or why not?

When switch Sin Fig. E25.29 is open, the voltmeter V reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, and the circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.

Fig. E25.29.

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.

Sign-up for free