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Solutions for College Physics, 11th edition by Hugh D Young

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Complete Solutions Manual for College Physics 11e 11th edition by Hugh D Young, Philip W. Adams. Full Chapters Solutions are included - Chapter 1 to 30 MECHANICS 1. Models, Measurements, and Vectors 2. Motion Along a Straight Line 3. Motion in a Plane 4. Newton's Laws of Motion 5. Applica...

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solutions
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9780136874300

Instelling College Physics
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Solutions for College Physics, 11th edition Young
All Chapters Solutions included Chap 1 - 30

MATHEMATICS REVIEW
0
Solutions to Problems

0.1. Set Up and Solve: (−3x 4 y 2 ) 2 = ( −3) 2 ( x 4 ) 2 ( y 2 ) 2 = 9 x8 y 4
Reflect: The square of any real expression must be positive.

(2344 ) 2 [23 (22 )4 ]2 (2328 )2 (211) 2 222
0.2. Set Up and Solve: = = = 2(22−12) = 210 = 1024
= =
(8) 4 (23 ) 4 212 212 212
Reflect: Since we are evaluating a numerical expression you could use your calculator to obtain the result without
any algebraic manipulation, but it is more precise in general to do the algebra first before using your calculator.

( )( )
2 2 2
 8 x3 y 2  82 x 3 y 2 64 x 6 y 4 16 x6
*0.3. Set Up and Solve:  5   = = =
( )
2 10
 2y  22 y 5 4y y6

Reflect: Notice that we could have used 8 = 23 in the numerator and canceled the two factors of 2 in the denominator,
leaving us with 24 = 16 in the numerator.

( )( )
5 5
 x −4 y −4 
5
x −4 y −4 x −20 y −20
0.4. Set Up and Solve:  − 2 −2  = ( −1)
5
= − 10 −10 = − x −30 y −10
( )( )
5 5
 x y  x 2 y −2 x y

Reflect: Any real expression raised to an odd power has the same sign as the original expression.

0.5. Set Up and Solve: The decimal point must be moved 5 places to the left to change 475000 into a number between
1 and 10. Thus, we have 475000 = 4.75 × 105.
Reflect: When written in scientific notation, numbers larger than 1 will have positive exponents and numbers less
than 1 will have negative exponents for their power of 10.

0.6. Set Up and Solve: The decimal point must be moved 6 places to the right to change 0.00000472 into a number
between 1 and 10. Thus, we have 0.00000472 = 4.72 × 10−6.
Reflect: When written in scientific notation, numbers larger than 1 will have positive exponents and numbers less
than 1 will have negative exponents for their power of 10.

*0.7. Set Up and Solve: The decimal point must be moved 2 places to the left to change 123 into a number
between 1 and 10. Thus, we have 123 × 10−6 = 1.23 × 102 × 10−6 = 1.23 × 10−4.

0-1

,0-2 Chapter 0

Reflect: Alternatively, we could have written our original number in decimal form by moving the decimal point
6 places to the left to obtain 123 × 10−6 = 0.000123. Finally, we could convert the result into scientific notation by
moving the decimal point 4 places to the right to obtain 0.000123 = 1.23 × 10−4.

8.3 × 105 8.3
0.8. Set Up and Solve: = × 10(5− 2) = 1.1 × 103 , where we have rounded the decimal number to the
7.8 × 102 7.8
nearest tenth.
Reflect: You can make a quick estimate, to check your result, by rounding each number to the nearest power of 10.
8.3 × 105 106
Thus, we have the estimate ≈ = 103 , which can be done without a calculator.
7.8 × 102 103

0.9. Set Up and Solve: First we subtract 9 x from both sides to obtain −5 x + 6 = −14. Next, we subtract 6 from
both sides to get −5 x = −20. Finally, we divide both sides of the equation by −5 to get the solution x = 4.
Reflect: There is no need to keep x on the left side of the equation. By adding −4 x + 14 to both sides of the
equation, we get 20 = 5 x, which reduces to 4 = x.

0.10. Set Up and Solve: Divide each side by m to obtain E m = c 2 , and then take the square root of each side to
obtain c = ± E m .

Reflect: Notice that we must retain both signs. The physics of the problem at hand will determine the sign.

*0.11. Set Up and Solve: Add −3x 2 − 6 to both sides to obtain x 2 = 12. Next, take the positive and negative
square root of both sides to obtain x = ± 12 = ± (3)(4) = ±2 3.
Reflect: Since this quadratic has no terms containing x, it is possible to solve it without factoring or using the
quadratic formula.

0.12. Set Up and Solve: Reflect the equation to obtain −9.8t 2 = −196. Next, divide both sides by −9.8 to
−196
obtain t 2 = = 20. Finally, take the positive and negative square root of both sides to obtain t = ± 20 =
−9.8
± (4)(5) = ±2 5.
Reflect: Since this quadratic has no terms containing t, it is possible to solve it without factoring or using the
quadratic formula.

*0.13. Set Up and Solve: Notice that −2 + −3 = −5 and (−2)(−3) = 6. Thus, we can factor the equation as x 2 − 5 x + 6
= ( x − 3)( x − 2) = 0. The two roots are x = 3 and x = 2.
−b ± b2 − 4ac
Reflect: Alternatively, we can use the quadratic formula with a = 1, b = −5, and c = 6 to obtain x = =
2a
5 ± (−5) 2 − 4(1)(6) 5 ± 1
= = 3 or 2.
2(1) 2

0.14. Set Up and Solve: The equation is satisfied if either factor is zero, which means that x − 5 = 0 or x + 3 = 0.
Solving these equations gives the two roots of the equation, x = 5 and x = −3.

Reflect: Alternatively, we can use the quadratic formula with a = 1, b = −2, and c = −15 to obtain
2 2
−b ± b − 4ac 2 ± (−2) − 4(1)(−15) 2 ± 8
x= = = = 5 or − 3.
2a 2(1) 2

, Mathematics Review 0-3

−b ± b 2 − 4ac
0.15. Set Up and Solve: Use the quadratic formula with a = 4.9, b = 2, and c = −20 to obtain t = =
2a
−2 ± 22 − 4(4.9)(−20) −2 ± 396
= = −2.2 or 1.8. The decimal answers are precise to the nearest tenth.
2(4.9) 9.8
Reflect: In physics, many numbers are based on approximate measurements. Therefore, it is usually okay to write
answers in approximate decimal form. For example, we may write 2 as 1.4 or 1.41, depending on the situation.

4y +1
0.16. Set Up and Solve: Solve the first equation for x to obtain x = and use this expression for x in the
5
 4y +1
second equation to obtain 6 y = 10   − 4 = 8 y − 2. Solving for y we obtain −2 y = −2 or y = 1. Thus, we have
 5 
4y +1 4 +1
x= = = 1. The solution is x = 1, y = 1.
5 5
Reflect: We could also solve this system by multiplying the first equation by 2 and adding the result to the second
equation.

*0.17. Set Up and Solve: Multiply the first equation by 2 and subtract 6 from each side to get 4 x = 10 y − 6. Insert
2 2
this value of 4x into the right-hand side of the second equation to get y − 1 = − (10 y − 6 ) or y − 1 = −10 y + 6.
3 3
Multiply both sides of the equation by 3 to get 2 y − 3 = −30 y + 18. Add 30y + 3 to each side to obtain 32 y = +21.
21
Finally, divide each side by 32 to get y = . To find x, insert this result into the equation 4 x = 10 y − 6 to get
32
21 210 192 18 9 9
4 x = 10 − 6 = − = = or, dividing each side by 4, x = .
32 32 32 32 16 64
Reflect: We could also solve this equation by solving the second equation for x and substituting the result into the
first equation.

x 2 1 8
0.18. Set Up and Solve: Since x ∝ y we have = or x = y. Substituting y = 8 we obtain x = = 1.6.
y 10 5 5
Reflect: In physics, variables often represent physical measurements and so it is common to represent fractions as
decimals.

0.19. Set Up and Solve: If F and m are directly proportional, we can write F = κ m, where κ is a constant of
proportionality to be determined. We are given that F = 9.8 when m = 1, so we can find κ :

F = κm
F 9.8 N
κ= = = 9.8 N/kg
m 1 kg

Using this constant, we can find the force when m = 2.8 :

F = κ m = ( 9.8 N/kg )( 2.8 kg ) = 27 N

Reflect: For this problem, the constant of proportionality κ tells us how many newtons of force are exerted per
unit mass.

, 0-4 Chapter 0

0.20. Set Up and Solve: Let T be the temperature of the gas (in degree kelvins) and V be its volume. Thus, we have
V 4. 0 L
V ∝ T , when the pressure is fixed, and we have = or V = (0.040 L/K)T . Thus, when T = 300 K we have
T 100 K
V = (0.040 L/K)T = (0.040 L/K)(300 K) = 12 L.
Reflect: Since the temperature increases by a factor of 3, the volume must also increase by a factor of 3.

1
*0.21. Set Up and Solve: Let A be the amplitude of the sound and d the distance from the source. Thus, A ∝
d
(4.8 × 10−6 m 2 )
and we have Ad = (4.8 × 10−6 m)(1.0 m) or A = . Thus, when d = 4.0 m we have A = 1.2 × 10−6 m.
d
Reflect: Since the distance is increased by a factor of 4, the amplitude is decreased by a factor of 4 (i.e., multiplied
1
by ).
4

κ
0.22. Set Up and Solve: If time is inversely proportional to speed, then we can write t = , where κ is a
υ
constant. We are given that it takes t1 = 1 h to travel 60 mi at speed υ1 = υ , and we want to find the time t2 it takes to
travel the same distance at speed υ 2 = υ / 3. We thus have two equations and two unknowns, κ and t2, so we can
solve for t2:

κ κ 
t1 = = t κ υ 1
υ1 υ  1
 = = or 3t1 = t2
κ κ 3κ  t2 υ 3κ 3
t2 = = =
υ 2 υ 3 υ 

Given that t1 = 1 h, then t2 = 3 × 1 h = 3 h.

Reflect: It makes sense that if we travel three times slower, then the travel time will be three times longer.

*0.23. Set Up and Solve: Let F be the force of gravity on an object that is a distance d from the earth’s center.
1
Thus, we have F ∝ or Fd 2 = k . Also, when d = 6.38 × 106 m we know that F = 700 N so we can find the
d2
value of k. When the astronaut is 6000 km (6 × 106 m) from the earth’s surface he is 6.38 × 106 + 6.00 × 106 =
k (700 N)(6.38 × 106 m) 2
1.238 × 107 m from the center of the earth. Thus, we have F = = = 186 N.
d2 (1.238 × 107 m) 2
2
F2  d1 
Reflect: We could also solve this problem using the ratio equation =   . Since the astronaut nearly doubles
F1  d 2 
his distance from the earth’s center as he moves from the surface into orbit, his weight decreases by nearly a factor of
1
4 (i.e., 2 ).
2

0.24. Set Up and Solve: From the plot of y versus xn, we see that a straight line is obtained for n = 3.

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