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# Exponents

Module by: Scott Starks. E-mail the author

Summary: This module is part of a collection of modules that present topics covered in a PreCalculus (MATH 1508) class at the University of Texas at El Paso.

## Exponents

### Introduction

Exponentiation is a mathematical operation that is employed extensively in applications in fields that range from science and engineering to finance and economics. This module will begin with a brief discussion of the terminology and the mathematics involved with exponentiation. This will be followed with some applications of exponentiation in science and engineering.

Scientific notation and engineering notation are introduced in this module as two common means for expressing physical quantities. Each of these representational schemes involve the use of exponents. Applications are presented on topics including electrical power, gravitatitational force and electrostatic force are presented as a means for illustrating how exponentiation can be useful in problem solving.

### Definition and Terminology

Exponentiation is an operation in mathematics that makes use of two numbers known as the base (x) and the exponent (n) and is expressed as

x n x n size 12{x rSup { size 8{n} } } {}
(1)

In this module, we will restrict our discussion to situations where the exponent (n) is an integer. Being an integer, the exponent may be positive, negative or equal to 0. We will consider each case below.

If the exponent n is a positive integer, then the operation of exponentiation is equivalent to the multiplication of the base x with itself a total of n times.

In the situation where the base (x) is 5 and the exponent (n) is 3, we obtain the result

5 3 = 5 5 5 = 125 5 3 = 5 5 5 = 125 size 12{5 rSup { size 8{3} } =5 cdot 5 cdot 5="125"} {}
(2)

Now, let us the case where the exponent is a negative integer. In this case, the process of exponentiation is equivalent to dividing one by the base x by n times.

Let us consider an example where the base (x) is 5, but now the exponent (n) is -3. In this example, we obtain the result

5 3 = 1 5 5 5 = 1 125 = 0 . 008 5 3 = 1 5 5 5 = 1 125 = 0 . 008 size 12{5 rSup { size 8{ - 3} } = { {1} over {5 cdot 5 cdot 5} } = { {1} over {"125"} } =0 "." "008"} {}
(3)

Suppose that we form the product of 5353 size 12{5 rSup { size 8{3} } } {} with 5353 size 12{5 rSup { size 8{ - 3} } } {}.

5 3 × 5 3 = 125 × 0 . 008 5 3 × 5 3 = 125 × 0 . 008 size 12{5 rSup { size 8{3} } times 5 rSup { size 8{ - 3} } ="125" times 0 "." "008"} {}
(4)
5 3 3 = 1 5 3 3 = 1 size 12{5 rSup { size 8{3 - 3} } =1} {}
(5)
5 0 = 1 5 0 = 1 size 12{5 rSup { size 8{0} } =1} {}
(6)

The result of the multiplication reminds us of the property of exponents that states whenever a base (x) is raised to an exponent (n) that is zero, the result is 1.

### Scientific and Engineering Notation

One of the most important uses of exponents in the fields of science and engineering is that of scientific notation. As is often the case, in the fields of science and engineering one often deals with numbers that are extremely large or extremely small. Scientific notation is an effective means for writing such number that makes use of exponents. In many cases, scientific notation enables one to write very large or very small numbers in a manner that is more convenient, insightful and compact that writing numbers using decimal notation.

Numbers can be written in scientific notation in the following form

a × 10 b a × 10 b size 12{a times "10" rSup { size 8{b} } } {}
(7)

where a is the coefficient which can be any real number in the range 1 < |a| <10 and b is an integer that represents the exponent.

In chemistry, the quantity of an element having a weight in grams numerically equal to that element’s atomic weight is the gram atomic weight of that element. This quantity is often refererred to as a gram atom. A gram atom of any element contains the same number of atoms as the gram atom of any other element. The number of atoms in any gram atom is called Avogadro’s number (N). Through meticulous experimental study, the value of Avogadro’s number has been determined as

N = 602 , 300 , 000 , 000 , 000 , 000 , 000 , 000 N = 602 , 300 , 000 , 000 , 000 , 000 , 000 , 000 size 12{N="602","300","000","000","000","000","000","000"} {}
(8)

Question: How can Avogadro’s number be expressed in scientific notation?

Solution: Avogadro’s number can be written as the product of a real number (6.023) by the term (100,000,000,000,000,000,000,000)

N = 6 . 023 × 100 , 000 , 000 , 000 , 000 , 000 , 000 , 000 N = 6 . 023 × 100 , 000 , 000 , 000 , 000 , 000 , 000 , 000 size 12{N=6 "." "023" times "100","000","000","000","000","000","000","000"} {}
(9)
N = 6 . 023 × 10 23 N = 6 . 023 × 10 23 size 12{N=6 "." "023" times "10" rSup { size 8{"23"} } } {}
(10)

Rather than trying to perform calculations using the decimal form for Avogadro’s number, chemists have learned to appreciate the more mathematically tractable scientific notation form of this important constant.

Engineering notation is quite similar to scientific notation. Numbers are written in engineering notation in the following form

a × 10 c a × 10 c size 12{a times "10" rSup { size 8{c} } } {}
(11)

where the coefficient a is a real number in the range 1 < |a |< 1,000 and the exponent c is restricted to be an integer multiple of 3.

An additional restriction on the coefficient (a) requires that a be expressed using no more than 3 significant digits. The restriction that the exponent be an integer multiple of 3 allows the numbers that result from the transformation to engineering notation to be expressed using the standard prefixes associated with the Scientifique Internationale (SI) system of units.

Example (Width of the Asteroid Belt)

Let’s consider the following application of engineering notation. The width of the asteroid belt has been determined to be 280,000,000 m.

Question: What is the width of the asteroid belt expressed in engineering notation?

Solution: Let us begin by expressing this quantity using scientific notation

280 , 000 , 000 m = 2 . 80 × 10 8 m 280 , 000 , 000 m = 2 . 80 × 10 8 m size 12{"280","000","000"m=2 "." "80" times "10" rSup { size 8{8} } m} {}
(12)

We notice that the exponent is not an integer multiple of 3, so this quantity is not yet expressed in engineering notation. We do know that

10 8 = 10 2 + 6 = 10 2 × 10 6 10 8 = 10 2 + 6 = 10 2 × 10 6 size 12{"10" rSup { size 8{8} } ="10" rSup { size 8{2+6} } ="10" rSup { size 8{2} } times "10" rSup { size 8{6} } } {}
(13)

This quantity can be substituted into the previous equation to yield the expression for the width of the asteroid belt in engineering notation

280 , 000 , 000 m = 2 . 80 × 10 8 m 280 , 000 , 000 m = 2 . 80 × 10 8 m size 12{"280","000","000"m=2 "." "80" times "10" rSup { size 8{8} } m} {}
(14)
280 , 000 , 000 m = 2 . 80 × 10 2 × 10 6 m 280 , 000 , 000 m = 2 . 80 × 10 2 × 10 6 m size 12{"280","000","000"m=2 "." "80" times "10" rSup { size 8{2} } times "10" rSup { size 8{6} } m} {}
(15)
280 , 000 , 000 m = ( 2 . 80 × 100 ) × 10 6 m 280 , 000 , 000 m = ( 2 . 80 × 100 ) × 10 6 m size 12{"280","000","000"m= $$2 "." "80" times "100"$$ times "10" rSup { size 8{6} } m} {}
(16)
280 , 000 , 000 m = 280 × 10 6 m 280 , 000 , 000 m = 280 × 10 6 m size 12{"280","000","000"m="280" times "10" rSup { size 8{6} } m} {}
(17)
280 , 000 , 000 m = 280 Mm 280 , 000 , 000 m = 280 Mm size 12{"280","000","000"m="280" ital "Mm"} {}
(18)

### Application: Electrical Power

We will begin our investigation of applications of exponents with a discussion of electrical power. Consider the electrical circuit diagram that shows a source voltage (V) attached to a resistor (R) to produce a current (I).

The relationship between V, R and I is summarized by Ohm’s Law

V = I × R V = I × R size 12{V=I times R} {}
(19)

where V is measured in volts (V) , I is measured in amps (A), and R is measured in ohms (Ω).

We may write an expression for the current as

I = V R I = V R size 12{I= { {V} over {R} } } {}
(20)

The power that is absorbed by the resistor is known to be the product of the current flowing through the resistor time the potential difference (voltage) across the terminals of the resistor. If we denote the power absorbed by the resistor as PR, then it can be expressed mathematically as

P R = I × V P R = I × V size 12{P rSub { size 8{R} } =I times V} {}
(21)

Substitution of the expression for I obtained from Ohm’s Law yields an equivalent expression for the power absorbed by the resistor

P R = V R V = V 2 R P R = V R V = V 2 R size 12{P rSub { size 8{R} } = left ( { {V} over {R} } right ) cdot V= { {V rSup { size 8{2} } } over {R} } } {}
(22)

Paying attention to the exponent, we can say that the power absorbed by the resistor is the square of the voltage across the terminals of the resistor divided by the resistance. The units associated with power are Watts (W).

We can obtain an alternative form for the power absorbed by the resistor. We begin with the general expression for the power absorbed by a resistor

P R = I × V P R = I × V size 12{P rSub { size 8{R} } =I times V} {}
(23)

If we now substitute the expression for V provided by Ohm’s Law into this equation, we obtain an alternative expression for power

P R = I ( I × R ) P R = I ( I × R ) size 12{P rSub { size 8{R} } =I cdot $$I times R$$ } {}
(24)
P R = I 2 R P R = I 2 R size 12{P rSub { size 8{R} } =I rSup { size 8{2} } R} {}
(25)

This relationship expresses the electrical power absorbed by the resistor as a product of the square of the current by the resistance.

Thus we have two equally useful formulas for the computation of the power absorbed by a resistor. One involves the square of the voltage, while the other incorporates the square of the current. Each formula involves a term that is raised to an exponent of 2.

Example (Electrical Power)

A 9 Volt battery is connected to a single 10 kΩ resistor.

Question: How much power is dissipated in the resistor? (Express the result in engineering notation.)

P R = ( 9 V ) 2 10 × 10 3 Ω P R = ( 9 V ) 2 10 × 10 3 Ω size 12{P rSub { size 8{R} } = { { $$9V$$ rSup { size 8{2} } } over {"10" times "10" rSup { size 8{3} } %OMEGA } } } {}
(26)
P R = 81 V 2 10 4 Ω P R = 81 V 2 10 4 Ω size 12{P rSub { size 8{R} } = { {"81"V rSup { size 8{2} } } over {"10" rSup { size 8{4} }  %OMEGA } } } {}
(27)
P R = 81 × 10 4 V 2 Ω P R = 81 × 10 4 V 2 Ω size 12{P rSub { size 8{R} } ="81" times "10" rSup { size 8{ - 4} } left ( { {V rSup { size 8{2} } } over { %OMEGA } } right )} {}
(28)
P R = 81 × ( 10 1 × 10 3 ) W P R = 81 × ( 10 1 × 10 3 ) W size 12{P rSub { size 8{R} } ="81" times $$"10" rSup { size 8{ - 1} } times "10" rSup { size 8{ - 3} }$$ W} {}
(29)
P R = ( 81 × 10 1 ) × 10 3 W P R = ( 81 × 10 1 ) × 10 3 W size 12{P rSub { size 8{R} } = $$"81" times "10" rSup { size 8{ - 1} }$$ times "10" rSup { size 8{ - 3} } W} {}
(30)
P R = 8 . 1 × 10 3 W P R = 8 . 1 × 10 3 W size 12{P rSub { size 8{R} } =8 "." 1 times "10" rSup { size 8{ - 3} } W} {}
(31)
P R = 8 . 1 mW P R = 8 . 1 mW size 12{P rSub { size 8{R} } =8 "." 1 ital "mW"} {}
(32)

### Gravitational Force

Every object of finite mass near the surface of the earth experiences a force that pulls it downward. This force is called weight. In the 17th Century, Sir Isaac Newton postulated that this force which attracts objects to the earth might well be a particular example of a more general kind of interaction between objects. Newton was engaged in the study of the motion of planets around the sun and felt that the planets should obey the same laws of motion as objects falling toward the earth due to their weight. In one of humankind’s greatest bits of insight, Newton discovered that the forces responsible for both planetary motion and the phenomenon of weight could be equally understood if one hypothesized that every object of finite mass attracts every other object of finite mass with a force that depends upon the individual masses of the objects. Such forces are known as gravitational forces and they serve as a basis for what is now known as Newton’s law of gravitation.

Let us consider two objects of respective masses m1 an m2 that are separated by a distance that we denote as r. Newton’s law of gravitation states that each object will exert on the other an attractive force, directed along a straight line connecting the two objects. The magnitude of the force that mass m1 exerts on m2 is equal to the magnitude of the force that m2 exerts on m1. The magnitude of the force that the first mass exerts on the second is provided by the equation

F = G m 1 m 2 r 2 . F = G m 1 m 2 r 2 . size 12{F=G { {m rSub { size 8{1} } m rSub { size 8{2} } } over {r rSup { size 8{2} } } }  "." } {}
(33)

Here, G represents the universal gravitational constant and is given by

G = 6 . 67 × 10 11 N m 2 / kg 2 . G = 6 . 67 × 10 11 N m 2 / kg 2 . size 12{G=6 "." "67" times "10" rSup { size 8{ - "11"} } N - m rSup { size 8{2} } / ital "kg" rSup { size 8{2} }  "." } {}
(34)

Because the universal constant is so small, the gravitational forces that exists between objects of ordinary mass are extremely small.

We can incorporate the value for G along with the use of exponents in the equation for the magnitude of the force to yield the expression that follows.

F = ( 6 . 67 × 10 11 Newton meter 2 / kg 2 ) ( m 1 m 2 ) r 2 F = ( 6 . 67 × 10 11 Newton meter 2 / kg 2 ) ( m 1 m 2 ) r 2 size 12{F= $$6 "." "67" times "10" rSup { size 8{ - "11"} }  ital "Newton" - ital "meter" rSup { size 8{2} } / ital "kg" rSup { size 8{2} }$$  $$m rSub { size 8{1} } m rSub { size 8{2} }$$ r rSup { size 8{ - 2} } } {}
(35)

Thus we can see that exponents play an important role in the formulation of Newton’s law of gravitation. Exponents are important to the expression of the universal gravitational constant and the attractive force between two objects depends upon the reciprocal of the square of the distance between the objects.

Example (Gravitational Force)

The mass of the sun is 2×1030kg2×1030kg size 12{2 times "10" rSup { size 8{"30"} }  ital "kg"} {}. The mass of the earth is 5.97×1024kg5.97×1024kg size 12{5 "." "97" times "10" rSup { size 8{"24"} } ital "kg"} {}. The approximate distance between the sun and the earth is 149.6×106km.149.6×106km. size 12{"149" "." 6 times "10" rSup { size 8{6} } ital "km" "." } {}

Question: What is the magnitude of the gravitational force that the sun exerts on the earth? (Express the result in scientific notation.)

Solution:

F = ( 6 . 67 × 10 11 N m 2 / kg 2 ) ( 2 × 10 30 kg ) ( 5 . 97 × 10 24 kg ) / ( 149 . 6 × 10 6 km ) 2 F = ( 6 . 67 × 10 11 N m 2 / kg 2 ) ( 2 × 10 30 kg ) ( 5 . 97 × 10 24 kg ) / ( 149 . 6 × 10 6 km ) 2 size 12{F= $$6 "." "67" times "10" rSup { size 8{ - "11"} } N - m rSup { size 8{2} } / ital "kg" rSup { size 8{2} }$$  $$2 times "10" rSup { size 8{"30"} } ital "kg"$$ $$5 "." "97" times "10" rSup { size 8{"24"} } ital "kg"$$ / $$"149" "." 6 times "10" rSup { size 8{6} } ital "km"$$ rSup { size 8{2} } } {}
(36)
F = ( 6 . 67 × 10 11 N m 2 / kg 2 ) ( 2 × 10 30 kg ) ( 5 . 97 × 10 24 kg ) / ( 149 . 6 × 10 9 m ) 2 F = ( 6 . 67 × 10 11 N m 2 / kg 2 ) ( 2 × 10 30 kg ) ( 5 . 97 × 10 24 kg ) / ( 149 . 6 × 10 9 m ) 2 size 12{F= $$6 "." "67" times "10" rSup { size 8{ - "11"} } N - m rSup { size 8{2} } / ital "kg" rSup { size 8{2} }$$  $$2 times "10" rSup { size 8{"30"} } ital "kg"$$ $$5 "." "97" times "10" rSup { size 8{"24"} } ital "kg"$$ / $$"149" "." 6 times "10" rSup { size 8{9} } m$$ rSup { size 8{2} } } {}
(37)
F = ( 6 . 67 × 10 11 N m 2 ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 149 . 6 × 10 9 m ) 2 F = ( 6 . 67 × 10 11 N m 2 ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 149 . 6 × 10 9 m ) 2 size 12{F= $$6 "." "67" times "10" rSup { size 8{ - "11"} } N - m rSup { size 8{2} }$$  $$2 times "10" rSup { size 8{"30"} }$$ $$5 "." "97" times "10" rSup { size 8{"24"} }$$ / $$"149" "." 6 times "10" rSup { size 8{9} } m$$ rSup { size 8{2} } } {}
(38)
F = ( 6 . 67 × 10 11 N m 2 ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 22 , 380 × 10 18 m 2 ) F = ( 6 . 67 × 10 11 N m 2 ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 22 , 380 × 10 18 m 2 ) size 12{F= $$6 "." "67" times "10" rSup { size 8{ - "11"} } N - m rSup { size 8{2} }$$  $$2 times "10" rSup { size 8{"30"} }$$ $$5 "." "97" times "10" rSup { size 8{"24"} }$$ / $$"22","380" times "10" rSup { size 8{"18"} } m rSup { size 8{2} }$$ } {}
(39)
F = ( 6 . 67 × 10 11 N ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 22 , 380 × 10 18 ) F = ( 6 . 67 × 10 11 N ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 22 , 380 × 10 18 ) size 12{F= $$6 "." "67" times "10" rSup { size 8{ - "11"} } N$$  $$2 times "10" rSup { size 8{"30"} }$$ $$5 "." "97" times "10" rSup { size 8{"24"} }$$ / $$"22","380" times "10" rSup { size 8{"18"} }$$ } {}
(40)
F = ( 6 . 67 × 2 × 5 . 97 ) × ( 10 11 × 10 30 × 10 24 ) N / ( 2 . 238 × 10 22 ) F = ( 6 . 67 × 2 × 5 . 97 ) × ( 10 11 × 10 30 × 10 24 ) N / ( 2 . 238 × 10 22 ) size 12{F= $$6 "." "67" times 2 times 5 "." "97"$$ times $$"10" rSup { size 8{ - "11"} } times "10" rSup { size 8{"30"} } times "10" rSup { size 8{"24"} }$$ N/ $$2 "." "238" times "10" rSup { size 8{"22"} }$$ } {}
(41)
F = ( 79 . 64 × 10 43 ) N / ( 2 . 238 × 10 22 ) F = ( 79 . 64 × 10 43 ) N / ( 2 . 238 × 10 22 ) size 12{F= $$"79" "." "64" times "10" rSup { size 8{"43"} }$$ N/ $$2 "." "238" times "10" rSup { size 8{"22"} }$$ } {}
(42)
F = ( 79 . 64 / 2 . 238 ) × 10 43 22 N F = ( 79 . 64 / 2 . 238 ) × 10 43 22 N size 12{F= $$"79" "." "64"/2 "." "238"$$ times "10" rSup { size 8{"43" - "22"} } N} {}
(43)
F = ( 35 . 6 ) × 10 21 N F = ( 35 . 6 ) × 10 21 N size 12{F= $$"35" "." 6$$ times "10" rSup { size 8{"21"} } N} {}
(44)
F = 3 . 56 × 10 22 N F = 3 . 56 × 10 22 N size 12{F=3 "." "56" times "10" rSup { size 8{"22"} } N} {}
(45)

### Coulomb’s Law

Electrons and protons are examples of charged particles. In the case of the proton, charge is positive. The charge of a proton is thus said to have positive polarity. On the other hand, an electron has a negative charge. Thus the charge of an electron is said to have negative polarity.

An electrostatic force exists between two charged particles. If the charges associated with the two particles are of the same polarity, the electrostatic force will be repulsive. On the other hand if the charges for two particles have opposite polarity, the electrostatic force will be attractive.

Through extensive laboratory work, the physicist Charles Coulomb first established a mathematical expression that calculates the magnitude of the electrostatic force that results from the interaction of two charged particles. This expression which is known as Coulomb’s law for electrostatic forces is given by

F = k q Q r 2 F = k q Q r 2 size 12{F=k left ( { {qQ} over {r rSup { size 8{2} } } } right )} {}
(46)

where q and Q are the values of the charges measured in the units Coulombs, r is the distance measured in meters that separates the charged particles and k is a constant

k = 9 × 10 9 Newton meters 2 / Coulomb 2 k = 9 × 10 9 Newton meters 2 / Coulomb 2 size 12{k=9 times "10" rSup { size 8{9} }  ital "Newton" - ital "meters" rSup { size 8{2} } / ital "Coulomb" rSup { size 8{2} } } {}
(47)

It is interesting to note the similarity of the general form of Coulomb’s law for electrostatic forces with Newton’s law of gravitational force. Just as was the case with the gravitational force, the magnitude of the electrostatic force decreases with the square of the distance separating the two particles.

Example (Electrostatic Force)

The charge on an electron is given by the equation

q e = 1 . 60 × 10 19 Coulombs ( C ) q e = 1 . 60 × 10 19 Coulombs ( C ) size 12{q rSub { size 8{e} } = - 1 "." "60" times "10" rSup { size 8{ - "19"} } ital "Coulombs" $$C$$ } {}
(48)

The charge associated with a proton is

q p =+ 1 . 60 × 10 19 Coulombs ( C ) q p =+ 1 . 60 × 10 19 Coulombs ( C ) size 12{q rSub { size 8{p} } "=+"1 "." "60" times "10" rSup { size 8{ - "19"} } ital "Coulombs" $$C$$ } {}
(49)

Suppose that an electron and a proton are separated by a distance of 106nanometers.106nanometers. size 12{"10" rSup { size 8{ - 6} } ital "nanometers" "." } {}

Questions: What is the magnitude of the electrostatic force that the protonexerts on the electron? Is the force attractive or repulsive?

Solution:

F = k q e q p r 2 F = k q e q p r 2 size 12{F=k { {q rSub { size 8{e} } q rSub { size 8{p} } } over {r rSup { size 8{2} } } } } {}
(50)
F = ( 9 × 10 9 N m 2 / C 2 ) ( 1 . 60 × 10 19 C ) ( 1 . 60 × 10 19 C ) ( 10 6 × nanometers ) 2 F = ( 9 × 10 9 N m 2 / C 2 ) ( 1 . 60 × 10 19 C ) ( 1 . 60 × 10 19 C ) ( 10 6 × nanometers ) 2 size 12{F= { { $$9 times "10" rSup { size 8{9} } N - m rSup { size 8{2} } /C rSup { size 8{2} }$$  $$- 1 "." "60" times "10" rSup { size 8{ - "19"} } C$$ $$1 "." "60" times "10" rSup { size 8{ - "19"} } C$$ } over { $$"10" rSup { size 8{ - 6} } times ital "nanometers"$$ rSup { size 8{2} } } } } {}
(51)
F = ( 9 × 10 9 ) ( 1 . 60 × 10 19 ) ( 1 . 60 × 10 19 ) N m 2 ( 10 6 × 10 9 m ) 2 F = ( 9 × 10 9 ) ( 1 . 60 × 10 19 ) ( 1 . 60 × 10 19 ) N m 2 ( 10 6 × 10 9 m ) 2 size 12{F= { { $$9 times "10" rSup { size 8{9} }$$  $$- 1 "." "60" times "10" rSup { size 8{ - "19"} }$$ $$1 "." "60" times "10" rSup { size 8{ - "19"} }$$ N - m rSup { size 8{2} } } over { $$"10" rSup { size 8{ - 6} } times "10" rSup { size 8{ - 9} } m$$ rSup { size 8{2} } } } } {}
(52)
F = ( 9 ) ( 1 . 60 ) ( 1 . 60 ) ( 10 9 × 10 19 × 10 19 ) N m ( 10 15 m ) 2 F = ( 9 ) ( 1 . 60 ) ( 1 . 60 ) ( 10 9 × 10 19 × 10 19 ) N m ( 10 15 m ) 2 size 12{F= { { $$9$$  $$- 1 "." "60"$$ $$1 "." "60"$$  $$"10" rSup { size 8{9} } times "10" rSup { size 8{ - "19"} } times "10" rSup { size 8{ - "19"} }$$ N - m} over { $$"10" rSup { size 8{ - "15"} } m$$ rSup { size 8{2} } } } } {}
(53)
F = ( 23 . 0 × 10 29 ) N ( 10 15 ) 2 F = ( 23 . 0 × 10 29 ) N ( 10 15 ) 2 size 12{F= { { $$"23" "." 0 times "10" rSup { size 8{ - "29"} }$$ N} over { $$"10" rSup { size 8{ - "15"} }$$ rSup { size 8{2} } } } } {}
(54)
F = 23 . 0 × 10 29 × 10 30 N F = 23 . 0 × 10 29 × 10 30 N size 12{F="23" "." 0 times "10" rSup { size 8{ - "29"} } times "10" rSup { size 8{"30"} } N} {}
(55)
F = 23 . 0 × 10 1 N F = 23 . 0 × 10 1 N size 12{F="23" "." 0 times "10" rSup { size 8{1} } N} {}
(56)
F = 230 N F = 230 N size 12{F="230"`N} {}
(57)

Because the charges on the electron and the proton differ in polarity, the electrostatic force is attractive.

### Exercises

1. The weight of the Space Shuttle is 4,470,000 lb. Express this weight in scientific notation.
2. The weight of a honey bee is 0.000 385 05 lb. Express this weight in scientific notation.
3. Perform the following multiplications. (a) (37.5 x 107) x (2.87 x 105), (b) (37.5 x 107) x (2.87 x 10-5).
4. Perform the following divisions. (a) (37.5 x 107) / (2.87 x 105), (b) (37.5 x 107) / (2.87 x 10-5).
5. The area of the United States is 9.83 x 106 km2. Its population is 3.10 x 106. What is the population density (people/km2) of the United States?
6. The distance between the Sun and the Earth is 1.47 x 1011 m. The speed of light is 299,792,458 m/s. How long does it take for light to travel between the Sun and the Earth?
7. The planned Blythe Solar Power Plant in California is expected to produce 968 MW. The Aswan Dam Power Plant in Egypt produces 2.1 gigawatts of power. How may solar power plants with similar power production as that of the Blythe Solar Power Plant would be needed to match the power production of the Aswan Dam Power Plant?
8. An electric circuit consists of a 9.00 V battery in series with a load that has a resistance of 13.78 kΩ. (a) Find the power delivered by the battery. (b) Find the power absorbed by the load.
9. The mass of the Earth is 5.97 x 1024 kg, while that of the moon is 7.36 x 1022 kg. The average distance between the Earth and the Moon is 384,403 km. What is he force exerted on the Moon by the Earth?
10. Two charges are separated by 4.66 x 10-8 m. Each has a positive charge of 3.6 x 10-18 Coulombs. What is the repulsive force that they exert on one another?

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#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

#### Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks