/*! 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} Problem 29 A current of \(5.00 \mathrm{~mA}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 29

A current of \(5.00 \mathrm{~mA}\) is enough to make your muscles twitch. Calculate how many electrons flow through your skin if you are exposed to such a current for \(10.0 \mathrm{~s}\).

Short Answer

Expert verified
Answer: Approximately \(3.13 \times 10^{17}\) electrons.

Step by step solution

01

Convert the current to Coulombs per second

Given current, \(I = 5.00 \mathrm{~mA}\). To convert milliamperes (mA) to Amperes (A), divide by 1000: \(I = \frac{5.00}{1000} \mathrm{~A} = 0.005 \mathrm{~A}\)
02

Calculate the total charge that passed through the skin

Given time, \(t = 10.0 \mathrm{~s}\). We can calculate the total charge that passed through the skin using the formula \(Q = I \times t\), \(Q = 0.005 \mathrm{~A} \times 10.0 \mathrm{~s} = 0.05 \mathrm{~C}\)
03

Find the number of electrons

The charge of a single electron is approximately \(1.6 \times 10^{-19} \mathrm{~C}\). To find the number of electrons, divide the total charge by the charge of a single electron: \(n = \frac{0.05 \mathrm{~C}}{1.6 \times 10^{-19} \mathrm{~C}}\) \(n \approx 3.13 \times 10^{17}\) electrons So, approximately \(3.13 \times 10^{17}\) electrons flow through the skin when exposed to such a current for 10.0 s.

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

Two protons placed near one another with no other objects close by would a) accelerate away from each other. b) remain motionless. c) accelerate toward each other. d) be pulled together at constant speed. e) move away from each other at constant speed.

A charge \(Q_{1}\) is positioned on the \(x\) -axis at \(x=a\). Where should a charge \(Q_{2}=-4 Q_{1}\) be placed to produce a net electrostatic force of zero on a third charge, \(Q_{3}=Q_{1}\), located at the origin? a) at the origin c) at \(x=-2 a\) b) at \(x=2 a\) d) at \(x=-a\)

A positive charge \(q_{1}=1.00 \mu \mathrm{C}\) is fixed at the origin, and a second charge \(q_{2}=-2.00 \mu \mathrm{C}\) is fixed at \(x=10.0 \mathrm{~cm} .\) Where along the \(x\) -axis should a third charge be positioned so that it experiences no force?

Two positive charges, each equal to \(Q\), are placed a distance \(2 d\) apart. A third charge, \(-0.2 Q\), is placed exactly halfway between the two positive charges and is displaced a distance \(x \ll d\) perpendicular to the line connecting the positive charges. What is the force on this charge? For \(x \ll d\), how can you approximate the motion of the negative charge?

A negative charge, \(-q\), is fixed at the coordinate (0,0) It is exerting an attractive force on a positive charge, \(+q,\) that is initially at coordinate \((x, 0)\). As a result, the positive charge accelerates toward the negative charge. Use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to show that when the positive charge moves a distance \(\delta \ll x\) closer to the negative charge, the force that the negative charge exerts on it increases by \(\Delta F=2 k q^{2} \delta / x^{3}\) .
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Kinematics Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Thermodynamics

Read Explanation

Electromagnetism

Read Explanation

Measurements

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.