College Physics for AP® Courses

# Chapter 11

College Physics for AP® CoursesChapter 11

## Problems & Exercises

1.

$1 . 610 cm 3 1 . 610 cm 3 size 12{1 "." "610""cm" rSup { size 8{3} } } {}$

3.

(a) 2.58 g

(b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath.

4.

$2 . 70 g/cm 3 2 . 70 g/cm 3 size 12{2 "." "70""g/cm" rSup { size 8{3} } } {}$

6.

(a) 0.163 m

(b) Equivalent to 19.4 gallons, which is reasonable

8.

$7 . 9 × 10 2 kg/m 3 7 . 9 × 10 2 kg/m 3 size 12{7 "." 9 times "10" rSup { size 8{2} } "kg/m" rSup { size 8{3} } } {}$

9.

$15.8 g/cm 3 15.8 g/cm 3 size 12{"15" "." 6"g/cm" rSup { size 8{3} } } {}$

10.

(a) $1018kg/m31018kg/m3 size 12{"10" rSup { size 8{"18"} } "kg/m" rSup { size 8{3} } } {}$

(b) $2×104m2×104m size 12{2 times "10" rSup { size 8{4} } m} {}$

11.

$3.59×106Pa3.59×106Pa$; or $521 lb/in 2 521 lb/in 2 size 12{"521""lb/in" rSup { size 8{2} } } {}$

13.

$2.36 × 10 3 N 2.36 × 10 3 N size 12{2 "." "36" times "10" rSup { size 8{3} } N} {}$

14.

0.760 m

16.

$hρ g units = ( m ) kg/m 3 m/s 2 = kg ⋅ m 2 / m 3 ⋅ s 2 = kg ⋅ m/s 2 1/m 2 = N/m 2 hρ g units = ( m ) kg/m 3 m/s 2 = kg ⋅ m 2 / m 3 ⋅ s 2 = kg ⋅ m/s 2 1/m 2 = N/m 2$

18.

(a) 20.5 mm Hg

(b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range

20.

$1 . 09 × 10 3 N/m 2 1 . 09 × 10 3 N/m 2 size 12{1 "." "09" times "10" rSup { size 8{3} } "N/m" rSup { size 8{2} } } {}$

22.

24.0 N

24.

$2.55×107Pa2.55×107Pa$; or 251 atm

26.

$5.76×103N5.76×103N size 12{5 "." "76" times "10" rSup { size 8{3} } N} {}$ extra force

28.

(a) $V = d i A i = d o A o ⇒ d o = d i A i A o . V = d i A i = d o A o ⇒ d o = d i A i A o . size 12{ V=d rSub { size 8{i} } A rSub { size 8{i} } =d rSub { size 8{o} } A rSub { size 8{o} } drarrow d rSub { size 8{o} } =d rSub { size 8{i} } left ( { {A rSub { size 8{i} } } over {A rSub { size 8{o} } } } right ) "." } {}$

Now, using equation:

$F 1 A 1 = F 2 A 2 ⇒ F o = F i A o A i . F 1 A 1 = F 2 A 2 ⇒ F o = F i A o A i . size 12{ { {F rSub { size 8{1} } } over {A rSub { size 8{1} } } } = { {F rSub { size 8{2} } } over {A rSub { size 8{2} } } } drarrow F rSub { size 8{o} } =F rSub { size 8{i} } left ( { {A rSub { size 8{o} } } over {A rSub { size 8{i} } } } right ) "." } {}$

Finally,

$W o = F o d o = F i A o A i d i A i A o = F i d i = W i . W o = F o d o = F i A o A i d i A i A o = F i d i = W i . size 12{W rSub { size 8{o} } =F rSub { size 8{o} } d rSub { size 8{o} } = left ( { {F rSub { size 8{i} } A rSub { size 8{o} } } over {A rSub { size 8{i} } } } right ) left ( { {d rSub { size 8{i} } A rSub { size 8{i} } } over {A rSub { size 8{o} } } } right )=F rSub { size 8{i} } d rSub { size 8{i} } =W rSub { size 8{i} } } {}$

In other words, the work output equals the work input.

(b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that $Wout=Win−WfWout=Win−Wf size 12{W rSub { size 8{"out"} } =W rSub { size 8{"in"} } - W rSub { size 8{f} } } {}$; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case.

29.

Balloon:

P g = 5.00 cm H 2 O, P abs = 1.035 × 10 3 cm H 2 O. P g = 5.00 cm H 2 O, P abs = 1.035 × 10 3 cm H 2 O. alignl { stack { size 12{P rSub { size 8{g} } =5 "." "00""cm"H rSub { size 8{2} } "O,"} {} # P rSub { size 8{"abs"} } =1 "." "035" times "10" rSup { size 8{3} } "cm"H rSub { size 8{2} } O "." {} } } {}

Jar:

P g = − 50.0 mm Hg , P abs = 710 mm Hg. P g = − 50.0 mm Hg , P abs = 710 mm Hg. alignl { stack { size 12{P rSub { size 8{g} } = - "50" "." 0"mm""Hg,"} {} # P rSub { size 8{"abs"} } ="710""mm""Hg" "." {} } } {}

31.

4.08 m

33.

ΔP = 38.7 mm Hg, Leg blood pressure = 159 119 . ΔP = 38.7 mm Hg, Leg blood pressure = 159 119 . alignl { stack { size 12{ΔP="38" "." 7"mm""Hg,"} {} # size 12{"Leg""blood""pressure"= { {"159"} over {"119"} } "." } {} } } {}

35.

$22 . 4 cm 2 22 . 4 cm 2 size 12{"22" "." 4"cm" rSup { size 8{2} } } {}$

36.

$91 . 7% 91 . 7% size 12{"91" "." 7%} {}$

38.

$815 kg /m 3 815 kg /m 3 size 12{"815""kg/m" rSup { size 8{3} } } {}$

40.

(a) 41.4 g

(b) $41.4cm341.4cm3 size 12{"41" "." 4"cm" rSup { size 8{3} } } {}$

(c) $1.09 g/cm31.09 g/cm3 size 12{1 "." "09""g/cm" rSup { size 8{3} } } {}$

42.

(a) 39.5 g

(b) $50cm350cm3 size 12{"50""cm" rSup { size 8{3} } } {}$

(c) $0.79g/cm30.79g/cm3 size 12{0 "." "79""g/cm" rSup { size 8{3} } } {}$

It is ethyl alcohol.

44.

8.21 N

46.

(a) $960kg/m3960kg/m3 size 12{"960" "kg/m" rSup { size 8{3} } } {}$

(b) $6.34%6.34%$

She indeed floats more in seawater.

48.

(a) $0.240.24 size 12{0 "." "24"} {}$

(b) $0.680.68 size 12{0 "." "68"} {}$

(c) Yes, the cork will float because $ρobj<ρethyl alcohol(0.678g/cm3<0.79g/cm3)ρobj<ρethyl alcohol(0.678g/cm3<0.79g/cm3) size 12{ρ rSub { size 8{"obj"} } <ρ rSub { size 8{"ethyl""alcohol"} } $$0 "." "678""g/cm" rSup { size 8{3} } <0 "." "79""g/cm" rSup { size 8{3} }$$ } {}$

50.

The difference is $0.006%.0.006%.$

52.

$F net = F 2 − F 1 = P 2 A − P 1 A = P 2 − P 1 A F net = F 2 − F 1 = P 2 A − P 1 A = P 2 − P 1 A size 12{F rSub { size 8{"net"} } =F rSub { size 8{2} } - F rSub { size 8{1} } =P rSub { size 8{2} } A - P rSub { size 8{1} } A= left (P rSub { size 8{2} } - P rSub { size 8{1} } right )A} {}$

$= h 2 ρ fl g − h 1 ρ fl g A = h 2 ρ fl g − h 1 ρ fl g A size 12{ {}= left (h rSub { size 8{2} } ρ rSub { size 8{"fl"} } g - h rSub { size 8{1} } ρ rSub { size 8{"fl"} } g right )A} {}$

$= h 2 − h 1 ρ fl gA = h 2 − h 1 ρ fl gA size 12{ {}= left (h rSub { size 8{2} } - h rSub { size 8{1} } right )ρ rSub { size 8{"fl"} } ital "gA"} {}$

where $ρflρfl size 12{ρ rSub { size 8{"fl"} } } {}$ = density of fluid. Therefore,

$F net = ( h 2 − h 1 ) Aρ fl g = V fl ρ fl g = m fl g = w fl F net = ( h 2 − h 1 ) Aρ fl g = V fl ρ fl g = m fl g = w fl size 12{F rSub { size 8{"net"} } = $$h rSub { size 8{2} } - h rSub { size 8{1} }$$ Aρ rSub { size 8{"fl"} } g=V rSub { size 8{"fl"} } ρ rSub { size 8{"fl"} } g=m rSub { size 8{"fl"} } g=w rSub { size 8{"fl"} } } {}$

where is $wflwfl size 12{w rSub { size 8{"fl"} } } {}$ the weight of the fluid displaced.

54.

$592 N/m 2 592 N/m 2 size 12{"592""N/m" rSup { size 8{2} } } {}$

56.

$2 . 23 × 10 − 2 mm Hg 2 . 23 × 10 − 2 mm Hg size 12{2 "." "23" times "10" rSup { size 8{ - 2} } "mm""Hg"} {}$

58.

(a) $1.65×10−3m1.65×10−3m size 12{1 "." "65" times "10" rSup { size 8{ - 3} } m} {}$

(b) $3.71×10–4m3.71×10–4m size 12{3 "." "71" times "10" rSup { size 8{4} } m} {}$

60.

$6 . 32 × 10 − 2 N/m 6 . 32 × 10 − 2 N/m size 12{6 "." "32" times "10" rSup { size 8{ - 2} } "N/m"} {}$

Based on the values in table, the fluid is probably glycerin.

62.

P w = 14 . 6 N/m 2 , P a = 4.46 N/m 2 , P sw = 7.40 N/m 2 . P w = 14 . 6 N/m 2 , P a = 4.46 N/m 2 , P sw = 7.40 N/m 2 . alignl { stack { size 12{P rSub { size 8{w} } ="14" "." 6"N/m" rSup { size 8{2} } , } {} # p rSub { size 8{a} } =4 "." "46""N/m" rSup { size 8{2} } , {} # P rSub { size 8{"sw"} } =7 "." "40""N/m" rSup { size 8{2} } "." {} } } {}

Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure.

64.

$5.1º 5.1º size 12{5 "." 1°} {}$

This is near the value of $θ=0ºθ=0º size 12{θ=0°} {}$ for most organic liquids.

66.

$− 2 . 78 − 2 . 78 size 12{ - 2 "." "78"} {}$

The ratio is negative because water is raised whereas mercury is lowered.

68.

479 N

70.

1.96 N

71.

$− 63.0 cm H 2 O − 63.0 cm H 2 O size 12{ - "63" "." 0"cm"H rSub { size 8{2} } O} {}$

73.

(a) $3.81×103N/m23.81×103N/m2 size 12{3 "." "81" times "10" rSup { size 8{3} } "N/m" rSup { size 8{2} } } {}$

(b) $28.7 mm Hg28.7 mm Hg size 12{"28" "." 7"mm""Hg"} {}$, which is sufficient to trigger micturition reflex

75.

(a) 13.6 m water

(b) 76.5 cm water

77.

(a) $3.98×106Pa3.98×106Pa size 12{3 "." "98" times "10" rSup { size 8{6} } "Pa"} {}$

(b) $2.1×10−3cm2.1×10−3cm size 12{2 "." 1 times "10" rSup { size 8{ - 3} } "cm"} {}$

79.

(a) 2.97 cm

(b) $3.39×10−6J3.39×10−6J size 12{3 "." "39" times "10" rSup { size 8{ - 6} } J} {}$

(c) Work is done by the surface tension force through an effective distance $h/2h/2 size 12{h/2} {}$ to raise the column of water.

81.

(a) $2.01×104N2.01×104N size 12{2 "." "01" times "10" rSup { size 8{4} } N} {}$

(b) $1.17×10−3m1.17×10−3m size 12{1 "." "17" times "10" rSup { size 8{ - 3} } m} {}$

(c) $2.56×1010N/m22.56×1010N/m2 size 12{2 "." "56" times "10" rSup { size 8{8} } "N/m" rSup { size 8{2} } } {}$

83.

(a) $1.38×104N1.38×104N size 12{1 "." "38" times "10" rSup { size 8{4} } N} {}$

(b) $2.81×107N/m22.81×107N/m2 size 12{2 "." "81" times "10" rSup { size 8{7} } N} {}$

(c) 283 N

85.

(a) 867 N

(b) This is too much force to exert with a hand pump.

(c) The assumed radius of the pump is too large; it would be nearly two inches in diameter—too large for a pump or even a primary cylinder. The pressure is reasonable for bicycle tires.

## Test Prep for AP® Courses

1.

(e)

3.

(a) 100 kg/m3 (b) 60% (c) yes; yes (76% will be submerged) (d) answers vary

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