/*! 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} Q40E Electric eels generate electric ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500V and produce currents of 80mA(or even larger). A typical pulse lasts for 10ms. What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

Short Answer

Expert verified

The power and energy delivered to the enemy is 40 W and 0.40 J respectively.

Step by step solution

01

Concept/Significance of steady current.

The term "steady current" refers to a current that does not fluctuate over time. Any stimulus generally responds with a transient, pulsing current that settles with time and has a steady-state current value.

02

Determination of power and energy delivered to the enemy with a single pulse

The power is given by,

P = V . l

Here, V is the maximum potential/pulse and I is the current given.

Substitute all the values in the above,

P= ( 500 V ) ( 80 mA )

=40 W

The energy delivered is given by,

E = Pt

Here, P is the power transferred to the enemy and t is the time.

Substitute all the values in the above,

E=(40W)(10×10-3s)=0.40J

Thus, the power and energy delivered to the enemy is 40 W and 0.40 J respectively.

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

A 140-g ball containing excess electrons is dropped into a 110-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.300 T and direction from east to west. If air resistance is negligibly small, find the magnitude ond direction of the force that this magnetic field exerts on the ball just as it enters the field.

(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

An electron moves at 1.40×106m/sthrough a regionin which there is a magnetic field of unspecified direction and magnitude 7.40×10-2T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle
between the electron velocity and the magnetic field?

When is a 1.5 - VAAA battery not actually a 1.5 - V battery? That is, when do this its terminals provide a potential difference of less than 1.5 V ?

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.