9 Chapter

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Chapter

9 I

Practice 9A, p. 324

Givens 1. Fg = 50.0 N apparent weight in water = 36.0 N rwater = 1.00 × 103 kg/m3

apparent weight in liquid = 41.0 N rmetal = 3.57 × 103 kg/m3

Solutions a. FB = Fg − apparent weight = 50.0 N − 36.0 N = 14.0 N Fg (50.0 N)(1.00 × 103 kg/m3) rmetal =  rwater =  FB 14.0 N rmetal = 3.57 × 103 kg/m3 b. FB = Fg − apparent weight = 50.0 N − 41.0 N = 9.0 N F (9.0 N)(3.57 × 103 kg/m3) rliquid = B rmetal =  Fg 50.0 N rliquid = 6.4 × 102 kg/m3

2. m = 2.8 kg

FB = Fg

l = 2.00 m

rwater Vg = (m + M)g

w = 0.500 m

M = rwater V − m = rwater (l wh) − m

h = 0.100 m

M = (1.00 × 103 kg/m3)(2.00 m)(0.500 m)(0.100 m) − 2.8 kg = 1.00 × 102 kg − 2.8 kg

rwater = 1.00 × 103 kg/m3

M = 97 kg

g = 9.81 m/s2

Copyright © by Holt, Rinehart and Winston. All rights reserved.

3. w = 4.0 m

l

= 6.0 m

h = 4.00 cm

Fg = FB mg = rwaterVg = rwater (wl h)g Fg = (1.00 × 103 kg/m3)(4.0 m)(6.0 m)(0.0400 m)(9.81 m/s2) = 9.4 × 103 N

rwater = 1.00 × 103 kg/m3 g = 9.81 m/s2 4. mballoon = 0.0120 kg rhelium = 0.179 kg/m3 r = 0.500 m rair = 1.29 kg/m3 g = 9.81 m/s2

a. FB = rairV g = rair 3pr 3 g 4

(1.29 kg/m3)(4p)(0.500 m)3(9.81 m/s2) FB =  = 6.63 N 3 b. mhelium = rheliumV = rhelium 3pr 3 4

(0.179 kg/m3)(4p)(0.500 m)3 mhelium =  = 0.0937 kg 3 Fg = (mballoon + mhelium )g = (0.0120 kg + 0.0937 kg)(9.81 m/s2) Fg = (0.1057 kg)(9.81 m/s2) = 1.04 N Fnet = FB − Fg = 6.63 N − 1.04 N = 5.59 N

Section One—Pupil’s Edition Solutions

I Ch. 9–1

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Section Review, p. 324

Givens

Solutions

3. mballoon = 650 kg

FB = rairVg

mpack = 4600 kg

I

Fg = (mballoon + mpack + mhelium )g 3

rair = 1.29 kg/m

mhelium = rheliumV 3

rhelium = 0.179 kg/m

FB = Fg rairVg =(mballoon + mpack + rheliumV )g mballoon + mpack 650 kg + 4600 kg  =  V=  rair − rhelium 1.29 kg/m3 − 0.179 kg/m3 5200 kg V = 3 = 4.7 × 103 m3 1.11 kg/m

4. a = 0.325 m/s2

Use Newton’s second law.

rsw = 1.025 × 10 kg/m

msa = FB − Fg = mswg − ms g

g = 9.81 m/s2

rsVa = rswVg − rsVg

3

3

rs(a + g) = rswg

g 9.81 m/s2 rs = rsw  = (1.025 × 103 kg/m3)  a+g 0.325 m/s2 + 9.81 m/s2

9.81 m/s2 rs = (1.025 × 103 kg/m3) 2 = 992 kg/m3 10.14 m/s

Practice 9B, p. 327

r2 = 15.0 cm F2 = 1.33 × 104 N

F F a. 1 = 2 A1 A2 FA F2 p r12 F2r12  =  F1 = 2 1 =  A2 p r22 r22 (1.33 × 104 N)(0.0500 m)2 F1 =  = 1.48 × 103 N (0.150 m)2 F F2 1.33 × 104 N   = 1.88 × 105 Pa = b. P = 2 =  A2 p r22 (p )(0.150 m)2

2. Fg = 1025 N w = 1.5 m

F Fg 1025 N P =  =  =  = 2.7 × 102 Pa A wl (1.5 m)(2.5 m)

l = 2.5 m 3. r = 0.40 cm

a. Pnet = Pb − Pt = 1.010 × 105 Pa − 0.998 × 105 Pa = 1.2 × 103 Pa

Pb = 1.010 × 105 Pa Pt = 0.998 × 105 Pa

b. Fnet = Pnet A = Pnet pr 2 Fnet = (1.2 × 103 Pa)(p)(4.0 × 10−3 m)2 = 6.0 × 10−2 N

I Ch. 9–2

Holt Physics Solution Manual


1. r1 = 5.00 cm

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Givens

Solutions

1. h = 11.0 km

P = P0 + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(11.0 × 103 m)

Po = 1.01 × 105 Pa

P = 1.01 × 105 Pa + 1.11 × 108 Pa = 1.11 × 108 Pa

r = 1.025 × 103 kg/m3

I

2

g = 9.81 m/s

a. P = Po + roil ghoil

2. h water = 20.0 cm hoil = 30.0 cm

P = 1.01 × 105 Pa + (0.70 × 103 kg/m3)(9.81 m/s2)(0.300 m)

roil = 0.70 × 103 kg/m3

P = 1.01 × 105 Pa + 2.1 × 103 Pa

rwater = 1.00 × 103 kg/m3

P = 1.03 × 105 Pa

g = 9.81 m/s2

b. Pnet = P + rwater gh water

Po = 1.01 × 105 Pa

Pnet = 1.03 × 105 Pa + (1.00 × 103 kg/m3)(9.81 m/s2)(0.200 m) Pnet = 1.03 × 105 Pa + 1.96 × 103 Pa = 1.05 × 105 Pa

3. Po = 0 Pa

P = Po + rgh 4

P = 2.7 × 10 Pa r = 13.6 × 103 kg/m3

P−P 2.7 × 104 Pa − 0 Pa h = o =  rg (13.6 × 103 kg/m3)(9.81 m/s2)

g = 9.81 m/s2

h = 0.20 m

4. P = 3 Po

P = P0 + rgh 5

Po = 1.01 × 10 Pa 3

3

r = 1.025 × 10 kg/m


g = 9.81 m/s2

P−P 3Po − Po 2Po  =  h = o =  rg rg rg (2)(1.01 × 105 Pa) h =  = 20.1 m (1.025 × 103 kg/m3)(9.81 m/s2)

Section Review, p. 331 1. Fg = 25 N w = 1.5 m

Fg = 15 N r = 1.0 m

Fg = 25 N w = 2.0 m

Fg = 25 N r = 1.0 m

F Fg a. P =  = 2 A w 25 N P = 2 = 11 Pa (1.5 m) Fg F b. P =  = 2 A pr 15 N P = 2 = 4.8 Pa (p)(1.0 m) F Fg c. P =  = 2 A w 25 N P = 2 = 6.2 Pa (2.0 m) F Fg d. P =  = 2 A pr 25 N P = 2 = 8.0 Pa (p)(1.0 m) a is the largest pressure


I Ch. 9–3

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Givens

Solutions

2. h = 366 m

P = rgh 3

3

r = 1.00 × 10 kg/m

P = (1.00 × 103 kg/m3)(9.81 m/s2)(366 m) = 3.59 × 106 Pa

g = 9.81 m/s2

I

4. T(°C) = 11°C

T(K) = T(°C) + 273 = 11°C + 273 = 284 K

5. h = 5.0 × 102 m

P = Po + rg h = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(5.0 × 102 m)

Po = 1.01 × 105 Pa

P = 1.01 × 105 Pa + 5.0 × 106 Pa = 5.1 × 106 Pa

r = 1.025 × 103 kg/m3

P 5.1 × 106 Pa N =  =  = 5.0 × 101 Po 1.01 × 105 Pa

g = 9.81 m/s2

Practice 9D, p. 337 1. h2 − h1 = 16 m flow rate = 2.5 × 10−3 m3/min g = 9.81 m/s2

1

1

a. P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 The top of the tank and the spigot are open to the atmosphere, so P1 = P2 = Po. If we assume that the hole is small, then v2 ≈ 0. 1

Po + 2 rv12 + rgh1 = Po + rgh2 1  rv 2 1 2

= rg(h2 − h1)

2

v1 = 2g(h2 − h1)

m/s v1 = 2g (h h )( 9. 81 2)(1 6m ) = 18 m/s 2 − 1) = (2 b. flow rate = Av1 flow rate 1 2 1  = A = pr 2 = p 2D = 4pD2 v1 (4)(2.5 × 10−3 m3/min)(1 min/60 s)  (p)(18 m/s)

4(flow rate)  = p v1

D = 1.7 × 10−3 m = 1.7 mm 2. r = 1.65 × 103 kg/m3 2

A1 = 10.0 cm

vi = 275 cm/s P1 = 1.20 × 105 Pa A2 = 2.50 cm2

a. A1 v1 = A2 v2 Av (10.0 cm2)(10−4 m2/cm2)(2.75 m/s) v2 = 11 =  = 11.0 m/s A2 2.50 × 10−4 m2 1

1

b. P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 Because h1 = h2, 1

P2 = P1 + 2 r(v12 − v22) 1

P2 = 1.20 × 105 Pa + 2 (1.65 × 103 kg/m3)[(2.75 m/s)2 − (11.0 m/s)2] 1

P2 = 1.20 × 105 Pa + 2 (1.65 × 103 kg/m3)(7.56 m2/s2 − 121 m2/s2) 1

P2 = 1.20 × 105 Pa − 2 (1.65 × 103 kg/m3)(113 m2/s2) P2 = 1.20 × 105 Pa − 0.932 × 105 Pa = 2.7 × 104 Pa

I Ch. 9–4



D=

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Solutions

3. v1 = 15 cm/s

1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2

v2 = 2 v1

The change in pressure, ∆P, is P2 − P1.

r = 1.29 kg/m3

Because h1 = h2, 1

1

I

1

P2 − P1 = 2 r (v12 − v22) = 2 r[v12 − (2v1)2] = 2 rv12 (1 − 4) P2 − P1 =

−3  2

(−3)(1.29 kg/m3)(0.15 m/s)2 rv12 =  2

P2 − P1 = −4.4 × 10 −2 Pa

Section Review, p. 337 2. P1 = 3.00 × 105 Pa 3

3

r = 1.00 × 10 kg/m v1 = 1.00 m/s 1

r2 = 4r1

a. A1v1 = A2 v2 Av p r 2v1 r12v1 v2 = 11 = 1 = = 16v1 1 2 A2 p r22 4r1 v2 = (16)(1.00 m/s) = 16.0 m/s 1

1

P1 + 2 rv12 +rgh1 = P2 + 2 rv22 +rgh2 Because h1 = h2 , 1

b. P2 = P1 + 2 r(v12 − v22) 1

P2 = 3.00 × 105 Pa + 2 (1.00 × 103 kg/m3)[(1.00 m/s)2 − (16.0 m/s)2] 1

P2 = 3.00 × 105 Pa + 2 (1.00 × 103 kg/m3)(1.00 m2/s2 − 256 m2/s2) 1

P2 = 3.00 × 105 Pa + 2 (1.00 × 103 kg/m3)(−255 m2/s2)


P2 = 3.00 × 105 Pa − 1.28 × 105 Pa = 1.72 × 105 Pa 6.0 cm 3. r1 =  = 3.0 cm 2 2.0 cm r2 =  = 1.0 cm 2 h2 − h1 = 2.00 m r = 1.00 × 103 kg/m3 V = 2.5 × 10−2 m3

a. flow rate = A2 v2 V  ∆ t flow rate V v2 =  =  =  A2 A2 pr22∆t 2.5 × 10−2 m3 v2 =  = 2.7 m/s (p)(0.010 m)2(30.0 s)

∆t = 30.0 s b. A1v1 = A2 v2 A v p r 2v2 r22v2 v1 = 22 = 2 =  r12 A1 p r12 (0.010 m)2(2.7 m/s) v1 =  = 0.30 m/s (0.030 m)2 1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 P1 − P2 = r 2 v22 − 2 v12 + g(h2 − h1) 1

1

P1 − P2 = (1.00 × 103 kg/m3 )2 (2.7 m/s)2 − 2 (0.30 m/s)2 + (9.81 m/s2)(2.00 m) 1

1

P1 − P2 = (1.00 × 103 kg/m3)(3.6 m2/s2 − 0.045 m2/s2 + 19.6 m2/s2) P1 − P2 = (1.00 × 103 kg/m3)(23.2 m2/s2) = 2.32 × 104 Pa


I Ch. 9–5

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Practice 9E, p. 341

Givens

Solutions

1. T1 = 27°C

T1 = (273 + 27)K = 3.00 × 102 K

3

V1 = 1.5 m

I

P1 = 0.20 × 105 Pa V2 = 0.70 m3 P2 = 0.80 × 105 Pa

P1V1 P2V2  =  T1 T2 5 3 P2V2T1 (0.80 × 10 Pa)(0.70 m )(3.00 × 102 K)  =  T2 =  = 5.6 × 102 K 5 (0.20 × 10 Pa)(1.5 m3) P1V1

2. P1 = 1.0 × 108 Pa

T1 = (273 + 15.0)K = 288 K

T1 = 15.0°C

T2 = (273 + 65.0)K = 338 K

N2 =

1 N 2 1

At constant volume:

T2 = 65.0°C

P1 P2  =  N1T1 N2T2  P1N2T2 P12N1T2 =  P2 =  T1N1 T1N1 1

(1.0 × 108 Pa)(338 K) PT P2 = 1 2 =  = 5.9 × 107 Pa (2)(288 K) 2T1 P2 = Po + rgh

h = 10.0 cm

P2 = 1.01 × 105 Pa + (13.6 × 103 kg/m3)(9.81 m/s2)(0.100 m)

rmerc = 13.6 × 103 kg/m3

P2 = 1.01 × 105 Pa + 0.133 × 105 Pa = 1.14 × 105 Pa

T2 = 27°C

T1 = (273 + 37)K = 3.10 × 102 K

T1 = 37°C

T2 = (273 + 27)K = 3.00 ×102 K

P1 = Po = 1.01 × 105 Pa 2

g = 9.81 m/s

P1V1 P2V2  =  T1 T2 P2V2T1 (1.14 × 105 Pa)(1.0 × 10−7m3)(3.10 × 102 K)  =  V1 =  = 1.2 × 10−7m3 T2 P1 (3.00 × 102 K)(1.01 × 105 Pa)

Section Review, p. 341 1

4. P2 = 2P1 T2 =

3 T 4 1

P1V1 P2V2  =  T1 T2 V2 P1T2 P14T1 6 3  =  = =  =  V1 T1P2 T 1P 4 2 1 2 1 3

3:2

I Ch. 9–6



3. V = 0.10 cm3

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Print Givens 5. P1 = 6.0 atm T 1 = 27°C

Solutions a. At constant volume: P1T2 = P2T1 Pressure triples; thus, P2 = 3P1. P1T2 = 3P1T1

I

3P T T2 = 11 P1 T1 = (273 + 27)K = 3.0 × 102 K T2 = 3T1 = (3)(3.0 × 102 K) = 9.0 × 102 K b. Pressure and volume double; thus, P2 = 2P1 and V2 = 2V1. P1V1 P2V2  =  T1 T2 P2V2T1 (2P1)(2V1)T1  =  T2 =  P1V1 P1V1 T2 = 4T1 = (4)(3.0 × 102 K) = 1.2 × 103 K

Chapter Review and Assess, pp. 343–349 8. Fg = 31.5 N apparent weight in water = 265 N rwater = 1.00 × 103 kg/m3


apparent weight in oil = 269 N ro = 6.3 × 103kg/m3

a. FB = Fg − apparent weight = 315 N − 265 N = 5.0 × 101 N Fg (315 N)(1.00 × 103 kg/m3) ro =  rwater =  FB 5.0 × 101 N ro = 6.3 × 103 kg/m3 b. FB = Fg − apparent weight = 315 N − 269 N = 46 N FB (46 N)(6.3 × 103 kg/m3) roil =  ro =  Fg 315 N roil = 9.2 × 102 kg/m3

9. Fg = 300.0 N apparent weight = 200.0 N ralcohol = 0.70 × 103 kg/m3

FB = Fg − apparent weight = 300.0 N − 200.0 N = 100.0 N Fg (300.0 N)(0.70 × 103 kg/m3) ro =  ralcohol =  FB 100.0 N ro = 2.1 × 103 kg/m3

16. P = 2.0 × 105 Pa

Fg = 4PA = (4)(2.0 × 105 Pa)(0.024 m2) = 1.9 × 104 N

A = 0.024 m2 17. P = 5.00 × 105 Pa 4.00 mm r =  = 2.00 mm 2

F = PA = P(pr 2) F = (5.00 × 105 Pa)(p)(2.00 × 10−3 m)2 = 6.28 N


I Ch. 9–7

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Givens

I

Solutions

0.64 cm 18. rA =  = 0.32 cm 2 3.8 cm rB =  = 1.9 cm 2 Fg,B = 500.0 N

FA Fg ,B  =  AA AB Fg ,B AA Fg,B(p rA2) Fg ,BrA2 FA =  =  =  AB p rB2 rB2 (500.0 N)(0.0032 m)2 FA =  = 14 N (0.019 m)2 F = 14 N downward

19. h = 2.50 × 102 m

a. P = Po + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(2.50 × 102 m)

rw = 1.025 × 103 kg/m3

P = 1.01 × 105 Pa + 2.51 × 106 Pa = 2.61 × 106 Pa

5

Po = 1.01 × 10 Pa g = 9.81 m/s2 30.0 cm r =  = 15.0 cm 2 23. h2 − h1 = 0.30 m g = 9.81 m/s2

b. F = PA = P(pr 2) = (2.61 × 106 Pa)(p)(0.150 m)2 = 1.84 × 105 N

1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 The top of the trough and the hole are both open to the atmosphere, so P1 = P2 = Po. Because the hole is small, we can assume that v2 ≈ 0. 1

Po + 2 rv12 + rgh1 = Po + rgh2 1  rv 2 2 1

= rg(h2 − h1 )

2

v1 = 2g(h2 − h1)

v1 = 2g m/s (h h )( 9. 81 2)(0 .3 0m ) = 2.4 m/s 2 − 1) = (2

A2 = 1.00 × 10−8 m2 F = 2.00 N

1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 Because the syringe is horizontal, the above equation simplifies as follows: 1

r = 1.00 × 103 kg/m3

1

P1 + 2 rv12 = P2 + 2 rv22 Also, P1 − P2 = P1 − P0 , which equals the gauge pressure in the barrel. F 2.00 N = 8.00 × 104 Pa P1,gauge = P1 − P2 =  =  A1 2.50 × 10−5 m2 Finally, assume v1 is negligible in comparison with the fluid speed inside the needle. 1

P1 − P2 = 2 rv22 v2 =

29. T1 = 325 K

2(P1 − P2)  = r

(2)(8.00 × 104 Pa)  = 12.6 m/s 1.00 × 103 kg/m3

At constant volume: 5

P1 = 1.22 × 10 Pa P2 = 1.78 × 105 Pa

P1 P2  =  T1 T2 P2 T1 (1.78 × 105 Pa)(325 K) T2 =  =  = 474 K P1 1.22 × 105 Pa

I Ch. 9–8



24. A1 = 2.50 × 10−5 m2

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Solutions

30. P1 = 7.09 × 104 Pa

T1 = (273 + 100.0)K = 373 K

T1 = 100.0°C

T2 = (273 + 0.0)K = 273 K

P2 = 5.19 × 104 Pa

At constant volume:

T2 = 0.0°C

P1 P2 P3  =  =  T1 T2 T3

3

P3 = 4.05 × 10 Pa

I

P3T1 (4.05 × 103 Pa)(373 K) T3 =  =  = 21.3 K P1 7.09 × 104 Pa

g

31. Fg = 4.5 N

Fg

r = 13.6 × 103 kg/m3

Fg m V =  =  =  gr r r 4.5 N V =  = 3.4 × 10−5 m3 (9.81 m/s2)(13.6 × 103 kg/m3)

g = 9.81 m/s2

F = PA = (1.01 × 105 Pa)(1.00 km2)(106 m2/km2) = 1.01 × 1011 Pa

32. A = 1.00 km2 P = 1.01 × 105 Pa

Fg = PA = P(4pr 2)

33. mm = 70.0 kg

Fg (mm + mc )g (70.0 kg + 5.0 kg)(9.81 m/s2)   P = 2 =  = 4p v 4p r 2 (4p )(0.010 m)2

mc = 5.0 kg r = 1.0 cm

(75.0 kg)(9.81 m/s2) P =  = 5.9 × 105 Pa (4p)(0.010 m)2

g = 9.81 m/s2


34. V1 = 8.20 × 10−4 m3

P2 = (0.95)(Po + rgh)

P1 = 0.95Po

P2 = (0.95)[(1.013 × 105 Pa) + (1.00 × 103 kg/m3)(9.81 m/s2)(10.0 m)]

h = 10.0 m

P2 = (0.95)(1.013 × 105 Pa + 0.981 × 105 Pa) = (0.95)(1.994 × 105 Pa) = 1.9 × 105 Pa

Po = 1.013 × 105 Pa 3

Using the ideal gas law, where T1 = T2:

3

r = 1.00 × 10 kg/m

P1V1 = P2V2

g = 9.81 m/s2

(0.95)(1.013 × 105 Pa)(8.20 × 10–4m3) PV V2 = 1 1 =  = 4.2 × 10–4 m3 P2 1.9 × 105 Pa

35. V1 = 1.50 cm3

P1 = Po + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(100.0 m) P1 = 1.01 × 105 Pa + 10.1 × 105 Pa = 11.1 × 105 Pa

h = 100.0 m T1 = T2

Using the ideal gas law, where P2 = Po and T1 = T2.

Po = 1.01 × 105 Pa

P1V1 = P2V2

g = 9.81 m/s2 3

3

r = 1.025 × 10 kg/m 36. r = 1.35 × 103 kg/m3 r = 6.00 cm

P1V1 (11.1 × 105 Pa)(1.50 cm3) V2 =  =  = 16.5 cm3 P2 1.01 × 105 Pa m = rV = r 23pr 3 = 3 rpr 3 14

2

(2)(1.35 × 103 kg/m3)(π)(6.00 × 10–2 m)3 m =  = 6.11 × 10–1kg 3


I Ch. 9–9

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Givens

Solutions

6.00 m 37. r =  = 3.00 m 2 h = 1.50 m

a. P = Po + rgh = 1.01 × 105 Pa + (1.00 × 103 kg/m3)(9.81 m/s2)(1.50 m) P = 1.01 × 105 Pa + 1.47 × 104 Pa = 1.16 × 105 Pa

Po = 1.01 × 105 Pa

I

r = 1.00 × 103 kg/m3 g = 9.81 m/s2

Fg mg (150 kg)(9.81 m/s2) b. P =  = 2 =  = 52 Pa A pr (p)(3.00 m)2

m = 150 kg 38. v1 = 30.0 m/s r = 1.29 kg/m3

1

1

a. P1 + 2 rv12 + rgh1 = P2 + 2 rv22+ rgh2 We assume that the difference in height between the two points is negligible, and note that v2 = 0. 1

1

P2 − P1 = 2 rv12 = 2 (1.29 kg/m3)(30.0 m/s)2 = 5.80 × 102 Pa A = 175 m2

b. Fnet = Pnet A = (5.80 × 102 Pa)(175 m2) Fnet = 1.02 × 105 N upward

39. r = 1050 kg/m3 h2 − h1 = 1.00 m g = 9.81 m/s2

40. r1 = 0.179 kg/m3

1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22+ rgh2 Assume v1 = v2. P1 − P2 = rg(h2 − h1 ) = (1050 kg/m3)(9.81 m/s2)(1.00 m) = 1.03 × 104 Pa T1 = (273 + 0.0)K = 273 K

T1 = 0.0°C

T2 = (273 + 100.0)K = 373 K

T2 = 100.0°C

At constant pressure:

VT V2 = 12 T1 m V =  , so r m2 m1T2  =   r2 r1T1 The amount of gas remains constant, so m1 = m2. rT (0.179 kg/m3)(273 K) r2 = 11 =  = 0.131 kg/m3 T2 373 K

41. Fg = 1.0 × 106 N

Fg = mg = rVg = rAhg

r = 1.025 × 103 kg/m3

Fg 1.0 × 106 N A =  =  rhg (1.025 × 103 kg/m3)(2.5 × 10–2m)(9.81 m/s2)

g = 9.81 m/s2

A = 4.0 × 103 m2

h = 2.5 cm

I Ch. 9–10 Holt Physics Solution Manual


V1 V2  =  T1 T2

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Solutions

42. r2 = 20.0 m

P1V1 P2V2  =  T1 T2

3

P2 = 3.0 × 10 Pa T2 = 200.0 K

P2V2T1  V1 =  T2 P1

5

P1 = 1.01 × 10 Pa

I

P2 3pr23T1 =  T2 P1 4

T1 = 300.0 K

4 pr 3 3 1

r1 =

r1 =

43. m = 1.0 kg + 2.0 kg = 3.0 kg rf = 916 kg/m3 mb = 2.0 kg 3

3

rb = 7.86 × 10 kg/m 2

g = 9.81 m/s

3 



4

3 T1  = pT2 P1

P pr 3 4 2 3 2

3

3

P2T1r23  T2P1

(3.0 × 103 Pa)(300.0 K)(20.0 m)3  = 7.1 m (200.0 K)(1.01 × 105 Pa)

rf rf For the spring scale, apparent weight of block = Fg,b − FB = Fg,b − Fg,b  = mbg 1 −  rb rb 916 kg/m3 apparent weight of block = (2.0 kg)(9.81 m/s2) 1 −  7.86 × 103 kg/m3 = (2.0 kg)(9.81 m/s2)(1 − 0.117)

apparent weight of block = (2.0 kg)(9.81 m/s2)(0.883) = 17 N For the lower scale, the measured weight equals the weight of the beaker and oil, plus a force equal to and opposite in direction to the buoyant force on the block. Therefore,

rf m r apparent weight = mg + FB = mg + Fg,b  = m + bf g rb rb

3

(2.0 kg)(916 kg/m ) (9.81 m/s2) apparent weight = 3.0 kg +  7.86 × 103 kg/m3 = (3.0 kg + 0.23 kg)(9.81 m/s2) Copyright © by Holt, Rinehart and Winston. All rights reserved.

apparent weight = (3.2 kg)(9.81 m/s2) = 31 N

44. rv = 600.0 kg/m3

FB = Fg,r

A = 5.7 m2

FB =rwater Vwater g = rwater(Ah)g

Vr = 0.60 m3

Fg,r = mrg = rr Vrg

rwater =1.0 × 103 kg/m3

rwater Ahg =rr Vr g

g = 9.81 m/s2

(600.0 kg/m3)(0.60 m3) r Vr =  = 6.3 × 10–2 m = 6.3 cm h = r  rwater A (1.0 × 103 kg/m3)(5.7 m2)

45. P1,gauge = 1.8 atm T1 = 293 K P2,gauge = 2.1 atm Po = 1.0 atm

a. P1 = P1,gauge + Po = 1.8 atm + 1.0 atm = 2.8 atm P2 = P2,gauge + Po = 2.1 atm + 1.0 atm = 3.1 atm At constant volume: P P 1 = 2 T1 T2 PT (3.1 atm)(293 K) T2 = 21 =  = 3.2 × 102 K P1 2.8 atm


I Ch. 9–11

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Givens

Solutions b. V1 = Vi At constant temperature: P1V1 = P2V2

I

P1 2.8 atm Vf = V2 = V i = Vi = 0.90Vi P2 3.1 atm ∆V = Vf − Vi = 0.90Vi − Vi = −0.10Vi 0.10Vi should be released from the tire 46.

l 1 = 4.0 m

At 220 m down:

3.0 m r =  = 1.5 m 2 h = 220 m

T1 = (273 + 25)K = 298 K

T1 = 25°C

T2 = (273 + 5.0)K = 278 K P2 = Po + rgh = 1.01 × 105 Pa + (1025 kg/m3)(9.81 m/s2)(220 m) P2 = 1.01 × 105 Pa + 2.2 × 106 Pa = 2.3 × 106 Pa

T2 = 5.0°C rsw = 1025 kg/m3 Po = 1.01 × 105 Pa g = 9.81 m/s2

P1V1 P2V2  =  T1 T2 P1V1T2  V2 =  T1P2 P1(pr 2 l 1)T2 pr 2l 2 =  T1P2

l2 =

(1.01 × 105 Pa)(4.0 m)(278 K) P1 l 1T2 =  = 0.16 m  (298 K)(2.3 × 106 Pa) T1P2

where l 2 is height of the remaining air inside the bell. hwater = l 1 − l 2 = 4.0 m − 0.16 m = 3.8 m 47. h = 26 cm

 F m g a. r =  =  =  g

y = 3.5 cm

V

Fg = 19 N 2

g = 9.81 m/s

hwy

hwyg

19 N r =  = 1.0 × 103 kg/m3 (0.26 m)(0.21 m)(0.035 m)(9.81 m/s2) Fg Fg 19 N b. P =  =  =  = 3.5 × 102 Pa A hw (0.26 m)(0.21 m) Fg Fg 19 N c. P =  =  =  = 2.1 × 103 Pa A hy (0.26 m)(0.035 m)

I Ch. 9–12



Fg

w = 21 cm

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Solutions

48. ∆y = −1.0 m

Use the equations for a horizontally-launched projectile to determine the water jet’s initial speed. ∆x vx =  ∆t 1 ∆y = Ny,i∆t − 2 g∆t 2

∆x = 0.60 m g = 9.81 m/s2

vy,i = 0 m/s, so

I

1

∆y = − 2 g∆t 2

g −2∆y

∆t =

∆x vx =  −2∆y  g

Use Bernoulli’s equation for the height of the tank, h, noting that P1 = P2 = Po, v2 ≈ 0, and v1 = vx . 1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 1

rg(h2 − h1) = 2 rv12

∆x2 −g∆x2    2 2∆y −∆x2 −2∆y 1 v1   h = h2 − h1 = 2  =  =  =  g 4∆y 2g g

2g 2

−(0.60 m) h =  = 9.0 × 10–2 m = 9.0 cm (4)(−1.0 m) 49. h1 = 5.00 cm h2 = 12.0 cm ∆x1 = ∆x2


g = 9.81 m/s2

Designate the position of the lower hole as point 1 and the position of the higher hole as point 2. Use the equations for horizontally-launched projectiles to obtain expressions for the initial speeds of the water streams. ∆x v1 = 1 ∆t1 ∆t1 =

−g = g 2∆y1

2h1

∆x1 ∆x v1 = 1 =  ∆t1 2h1  g

Similarly, ∆x2 ∆x v2 = 2 =  ∆t2 2h2  g ∆x1 = ∆x2, so

v1 ∆t1 = v2∆t2

g h  =v h 2h g 2h2

∆t v1 = v2 2 = v2 ∆t1

1

2

2

1


I Ch. 9–13

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Givens

Solutions Solve for v2. Apply Bernoulli’s equation, with P1 = P2 = Po. 1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 1  r 2

I

1  2

(v12 − v22) = rg(h2 − h1)

v22h2  − v22 = g (h2 − h1) h1

2g(h2 − h1) 2gh1(h2 − h1)  = 2gh1 v22 =  =  (h2 − h1) h2  − 1 h1

v2 = 2g h1

Apply Bernoulli’s equation again, using h3 to represent the height of the tank. Note that P3 = P2 = Po, and v3 ≈ 0. 1

1

P2 + 2 rv22 + rgh2 = P3 + 2 rv32 + rgh3 1  rv 2 2 2

= rg(h3 − h2) v22 2gh1 h3 − h2 =  =  = h1 2g 2g h3 = h1 + h2 = 5.00 cm + 12.0 cm = 17.0 cm 50. Am = 6.40 cm2

F F 1 = 2 A1 A2

Ab = 1.75 cm2 mk = 0.50

Ab 1.75 cm2 2 (44 N) = 12 N Fb = F = p Am 6.40 cm

Fp = 44 N

Fb is the normal force exerted on the brake shoe. Fk is given as follows: Fk = mkFn = (0.50)(12 N) = 6.0 N

flow rate = 1.55 m3/s 52. h = 2.0 cm = 0.020 m y = 1.5 cm = 0.015 m

flow rate = Av = pr 2v 1.55 m3/s flow rate   = 31.6 m/s v =  = (p )(0.125 m)2 pr 2 Before the oil is added: Fg,b = FB,water = rwaterVg = rwater Ayg

3

roil = 900.0 kg/m

rwater = 1.00 × 103 kg/m3

After the oil is added: Fg,b = FB,water + FB,oil rwater Ayg = rwater A(h − y1)g + roil Ay1g rwatery = rwater h − rwater y1 + roil y1

1 − r  y = h − y roil

1

water

0.020 m − 0.015m h−y y1 =  =  900.0 kg/m3 roil  1 −  1−  1.00 × 103 kg/m3 rwater

0.0050 m 0.0050 m y1 =  =  = 5.0 × 10−2 m = 5.0 cm 1 − 0.9000 0.1000

I Ch. 9–14



0.250 m 51. r =  = 0.125 m 2

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Solutions

53. roil = 930 kg/m3

FB = Fg

h = 4.00 cm

FB,oil + FB,water = Fg,b

rb = 960 kg/m3

moil g + mwater g = mb g

rwater = 1.00 × 103 kg/m3

roilVoil + rwaterVwater = rbVb

g = 9.81 m/s2

roil A(h − y) + rwater Ay = rb Ah

I

roil (h − y) + rwater y = rbh roil h − roil y + rwater y = rb h y(rwater − roil ) = h(rb − roil ) h( rb − roil ) (0.0400 m)(960 kg/m3 − 930 kg/m3)  =  y= rwater − roil 1.00 × 103 kg/m3 − 930 kg/m3 (0.0400 m)(30 kg/m3) y =  = 1.71 × 10−2 m = 1.71 cm 70 kg/m3 54. Fg,b = 50.0 N apparent weight of sinker = 200.0 N − Fg,b

FB,b = Fg,b − (apparent weight of block and sinker − apparent weight of sinker) FB,b = Fg,b − [140.0 N − (200.0 N − Fg,b)]

apparent weight of block and sinker = 140.0 N

FB,b = Fg,b + 60.0 N − Fg,b = 60.0 N

rwater = 1.00 × 103 kg/m3

Fg,b (50.0 N)(1.00 × 103 kg/m3) rb =  rwater =  = 833 kg/m3 FB,b 60.0 N

55. ∆t = 1.0 s

For one molecule:

A = 8.0 cm2

mvf − mvi F ∆p P =  =  =  A A ∆t A ∆t

m = 4.68 × 10−26 kg

In a perfect elastic collision with the wall, vi = vf .

v i = 300.0 m/s

mvi − m(−vi) mvi + mvi 2mv P =  =  = i A∆t A∆t A∆t

N = 5.0 × 10


FB,b = Fg,b − apparent weight of blocks

23

For all of the molecules: 2mv (5.0 × 1023)(2)(4.68 × 10−26 kg)(300.0 m/s) P = N i =  A∆t (8.0 cm2)(1 × 10−4 m2/cm2)(1.0 s)

P = 1.8 × 104 Pa


I Ch. 9–15

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Givens

Solutions

56. h = 10.0 m

l y = l (sin q ) h2 − h1 = h − l y = h − l (sin q)

l

= 2.0 m

q = 30.0°

h2 − h1 = 10.0 m − (2.0 m)(sin 30.0°) = 10.0 m − 1.0 m = 9.0 m 2

I

g = 9.81 m/s

Apply Bernoulli’s equation to find v1. Assume that P1 = P2 = Po , v2 ≈ 0, and r is constant. 1

1

P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2 1 2  v 2 1

+ gh1 = gh2

v1 = 2g m/s m) = 13 m/s (h 2−h )( 9. 81 2)(9 .0 1) = (2 Find the height of the water when vy,f = 0 m/s by using an equation for projectile motion. vy,f 2 = vi2(sin q )2 − 2g∆y = 0

2

vi(sin q )  = ∆y =  2g 57. r2 = 2.0 mm

= 2.2 m above the spout opening

a. At constant temperature:

r1 = 3.0 mm

P1V1 = P2V2 3

3

r = 1.025 × 10 kg/m 5

P1 = Po = 1.01 × 10 Pa g = 9.81 m/s2

2

(13 m/s)(sin 30.0°)  (2)(9.81 m/s2)

3  P1V1 P13pr1 P1r13 P2 =  =  =  4 V2 r23 pr 3 3 2 4

P2 = P1 + rgh

r3 P1 13 − 1 r2 P2 − P1  =  h = rg rg

(3.0 × 10−3 m)3 −1 (1.01 × 105 Pa)  (2.0 × 10−3 m)3  h= (1.025 × 103 kg/m3)(9.81 m/s2)

(1.01 × 105 Pa)(2.4) (1.01 × 105 Pa)(3.4 − 1)  h =  = (1.025 × 103 kg/m3)(9.81 m/s2) (1.025 × 103 kg/m3)(9.81 m/s2) h = 24 m

b. P = Po + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(24 m) P = 1.01 × 105 Pa + 2.4 × 105 Pa = 3.4 × 105 Pa or, alternatively P1 r13 (1.01 × 105 Pa)(3.0 × 10–3m)3 P2 =  = 3.4 × 105 Pa 3 =  r2 (2.0 × 10–3 m)3

I Ch. 9–16



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Solutions

58. r1 = 0.30 m

flow rate1 flow rate1 0.20 m3/s v1 =  =  = 2 = 0.71 m/s 2 A1 πr1 (π)(0.30 m)

flow rate1 = 0.20 m3/s r2 = 0.15 m

A1v1 flow rate1 0.20 m3/s v2 =  =  = 2 = 2.8 m/s 2 A2 πv2 (π)(0.15 m)

h1 − h2 = 0.60 m

Apply Bernoulli’s equation to find the gauge pressure (P2 − P1) in the lower pipe.

g = 9.81 m/s2

P1 + 2 rv12 + rgh1 = P2 + 2 r v22 + rgh2P2 − P1 = 2 r(v12 − v22) + rg(h1 − h2) 1 = r 2 (v12 − v22) + g(h1 − h2)

P1 = Po

rwater = 1.00 × 103 kg/m3

1

1

I

1

P2 − P1 = (1.00 × 103 kg/m3)2 (0.71 m/s)2 − 2 (2.8 m/s)2 + (9.81 m/s2)(0.60 m)] 1

1

P2 − P1 = (1.00 × 103 kg/m3)(0.25 m2/s2 − 3.9 m2/s2 + 5.9 m2/s2) P2 − P1 = (1.00 × 103 kg/m3)(2.2 m2/s2) P2 − P1 = 2.2 × 103 Pa 59. k = 90.0 N/m

Fnet = FB − Fg,b − Fg,hel − Fspring = 0

mb = 2.00 g

rairVg − mb g − rhelVg − k∆x = 0

V = 5.00 m3

g(rairV − mb − rhelV) ∆x =  k

g = 9.81 m/s2 rair = 1.29 kg/m3 rhel = 0.179 kg/m3

(9.81 m/s2)[(1.29 kg/m3)(5.00 m3) − 2.00 × 10−3 kg − (0.179 kg/m3)(5.00 m3)] ∆x =  90.0 N/m (9.81 m/s2)(6.45 kg − 2.00 × 10−3 kg − 0.895 kg) ∆x =  90.0 N/m (9.81 m/s2)(5.55 kg) ∆x =  = 0.605 m 90.0 N/m


60. A = 2.0 cm2

a. flow rate = Av = (2.0 cm2)(42 cm/s) = 84 cm3/s

r = 1.0 g/cm3

In g/s:

v = 42 cm/s

flow rate = (84 cm3/s)(1.0 g/cm3) = 84 g/s

A2 = 3.0 × 103 cm2

b. Use the continuity equation. Av (2.0 cm2)(42 cm/s) v2 = 11 =  = 0.028 cm/s = 2.8 × 10−4 m/s A2 3.0 × 103 cm2

1.6 cm 61. ra =  = 0.80 cm 2 1.0 × 10−6 m rc =  2 = 0.50 × 10−6 m va = 1.0 m/s vc = 1.0 cm/s

Use the continuity equation. A va Ac = a, vc

where Ac is the total capillary cross section needed.

(p)(8.0 × 10−3 m)2(1.0 m/s) pr 2v Ac = aa =  = 2.0 × 10−2 m2 vc 0.010 m/s Ac = NA 2.0 × 10−3 m2 A Ac  = = 2.6 × 1010 capillaries N = c =  A p rc2 (p )(0.50 × 10−6 m)2


I Ch. 9–17

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Givens 62.

l

Solutions

= 1.5 m

V V l wh ∆t =  =  =  flow rate Av pr 2v

w = 65 cm h = 45 cm

I

(1.5 m)(0.65 m)(0.45 m) ∆t =  = 9.3 × 102 s (p )(0.010 m)2(1.5 m/s)

2.0 cm r =  = 1.0 cm 2 v = 1.5 m/s

Fnet = ma = FB − Fg

63. rair = 1.29 kg/m3

FB − Fg rairVg − rhelVg a =  =  rhelV m

rhel = 0.179 kg/m3 g = 9.81 m/s2

g(rair − rhel) r  = g air − 1 a= rhel rhel

1.29 kg/m3 a = (9.81 m/s2) 3 − 1 = (9.81 m/s2)(7.21 − 1) 0.179 kg/m a = (9.81 m/s2)(6.21) = 60.9 m/s2 Fnet = (m + mair)a = FB − Fg,b − Fg,a

64. m = 1.0 kg

(m + mair)a = rwaterVg − g(m + mair)

r = 0.10 m h = 2.0 m 3

rwater = 1.00 × 10 kg/m g = 9.81 m/s2 rair = 1.29 kg/m3

rwater 3pr 3 g rwaterVg rwater(4pr3)g   −g a =  − g =  − g = 4 m + mair 3m + rair (4pr3) m + rair 3pr 3 4

3

(1.00 × 103 kg/m3)(4p)(0.10 m)3(9.81 m/s2) a =  − 9.81 m/s2 (3)(1.0 kg) + (1.29 kg/m3)(4p)(0.10 m)3

(1.00 × 103 kg/m3)(4π)(0.10 m3)(9.81 m/s2) a =  − 9.81 m/s2 3.0 kg a = 41 m/s2 − 9.81 m/s2 = 31 m/s2 Use the following equation to find the speed of the ball as it exits the water. Note that vi = 0. vf 2 = vi2 + 2ah = 2ah Use the following equation to find the maximum height of the ball above the water. Note that vi = vf for the ball leaving the water. vf 2 = vi2 − 2g∆y = 0 v 2 2ah ah ∆y = i =  =  2g 2g g (31 m/s2)(2.0 m) ∆y =  = 6.3 m 9.81 m/s2

I Ch. 9–18



(1.00 × 103 kg/m3)(4p)(0.10 m)3(9.81 m/s2) a =  − 9.81 m/s2 3.0 kg + 0.016 kg

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Solutions

65. r = 0.60rwater 3

3

rwater = 1.00 × 10 kg/m ∆y = −10.0 m

First find the speed of the sphere just before impact by using the following equation. Assume vi = 0. vf 2 = vi2 − 2g∆y = −2g∆y

m) = 14.0 m/s vf = − 2g ∆ y = (− 2) (9 .8 1m /s 2)(− 10 .0

g = 9.81 m/s2

I

This is the initial velocity of the sphere as it enters the water. Now find the net force on the sphere to determine its acceleration underwater. Fnet = ma = FB − Fg 0.60rwaterVa = rwaterVg − 0.60rwaterVg 2 2 g − 0.60g 9.81 m/s − (0.60)(9.81 m/s ) a =  =  0.60 0.60

9.81 m/s2 − 5.9 m/s2 3.9 m/s2 a =  =  = 6.5 m/s2 0.60 0.60 Use the following equation to find the maximum depth: vf 2 = vi2 − 2ah = 0 (14.0 m/s)2 v2 h = i =  = 15 m 2a (2)(6.5 m/s2) 66. T1 = 27°C

T1 = (273 + 27) K = 3.00 × 102 K 5

P1 = 1.01 × 10 Pa

T2 = (273 + 225) K = 498 K

T2 = 225°C

At constant volume: P1 P2  =  T1 T2 P1 T2 (1.01 × 105 Pa)(498 K) P2 =  =  = 1.68 × 105 Pa T1 3.00 × 102 K


67. ∆t = 1.00 min N = 150 m = 8.0 g v = 400.0 m/s A = 0.75 m2

For one bullet: mvf − mvi F ∆p P =  =  =  A ∆t A A ∆t In a perfect elastic collision with the wall, vi = − vf. 2m v P =  A ∆t For all the bullets: 2m v (150)(2)(8.0 × 10−3 kg)(400.0 m/s) P = N  =  = 21 Pa A ∆t (0.75 m2)(1.00 min)(60 s/min)


I Ch. 9–19

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Givens

Solutions

68. m = 4.00 kg

a. Assume that the mass of the helium in the sphere is not significant compared with the 4.00 kg mass of the sphere.

0.200 m r =  = 0.100 m 2

I

Fnet = ma = FB − Fg ma = rwaterVg − mg

h = 4.00 m 3

3

rwater = 1.00 × 10 kg/m g = 9.81 m/s2

rwater 3pr 3g − mg 4

a=

m

(1.00 × 103 kg/m3)3(p)(0.100 m)3(9.81 m/s2) − (4.00 kg)(9.81 m/s2) a =  4.00 kg 4

41.1 N − 39.2 N 1.9 N a =  =  = 0.48 m/s2 4.00 kg 4.00 kg b. Noting that vi = 0, 1

1

∆y = vi ∆t + 2a∆t 2 = 2a∆t2

2∆ay = 2( ha− 2r) (2)(3.80 m) (2)(4.00 m − 0.200 m)  =  ∆t = = 4.0 s 0.48 m/ s 0.48 m/s ∆t =

2

2

Fnet = FB − Fg − Fspring = 0

69. k = 16.0 N/m mb = 5.00 × 10

rwaterVg − mb g − k∆x = 0

kg

3

rb = 650.0 kg/m

3

3

rwater = 1.00 × 10 kg/m g = 9.81 m/s2

m rwater b g − mb g − k∆x = 0 rb

m rwater b g − mb g rb ∆x =  k  − 1 m g r ∆x =  rwater

b

b

k

 − 1 (5.00 × 10 kg)(9.81 m/s ) 650.0 kg/m ∆x =  1.00 × 103 kg/m3

−3

2

3

16.0 N/m

(1.54 − 1)(5.00 × 10–3kg)(9.81 m/s2) ∆x =  16.0 N/m (0.54)(5.00 × 10−3 kg)(9.81 m/s2) ∆x =  = 1.7 × 10−3 m 16.0 N/m

I Ch. 9–20



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