exercise10_sol

# exercise10_sol - 251 College Physics I Exercise 10...

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Unformatted text preview: 251 College Physics I Exercise 10 Suggested solutions 1. After the two vessels are cormected. the equilibrium height of the liquid is: : (’21 + he) 2 In equalizing the levels of the two vessels. :1 liquid column of height (hl — h) falls by a height of (h —- I12). So the work done by the gravitational force is equal to: . l o W = mgAh = psi-4011 - h)(h - ’12) = ZPQAihi - he)" [2 2. (a) The ﬂuid in the needle is conﬁned. and so Pascal‘s principle may be applied. Pplungar = Pneadle Fplunuer F needle 3 : Aplunger Aneedle A 1‘2 => Fneedle = Ii-inlungei'Auéwdle = phmgerjﬁ Pluntl'! ’ plunger Hence, ..2 -3 2 Fm... = Fpluncer it?“ = (2.4 N)(0'10 X 10 m) = 5.68 x 10-4N 7 plunge, (0.65 x 10—2111)2 (1)) The force on the plunger is equal to: 133N/mr" 740.65 x10‘rgm)2 = 0.318 N l nun—Hg Fplumzer = PplungexAphmm = (lSmm-Hg) ( 3. The downward force 13:10..“ acting on the beaker is due to the weight of the beaker and the water inside the beaker. On the other hand, the upward force Fup acting on the beaker is due to the buoyant force exerted by the surrounding water. Therefore, V de = Pm: + 17159 771 Fup = png + Vb) = ng (V + j) where p“, density of water. L7 mi, 2 mass of the beaker, volume of the space inside the beaker. V2, 2 volume of the beaker. pr, 2 density of the material made up the beaker. As the beaker ﬂoats in the water, it must be true that: Fdown = Fup V => P109 + "log : pwg + 2 p5 => Pu:pr + 2mbe = 2% (PbV +7710) 2pwmb 27715 — pwV =>Pb= Hence, the density of the material made up the beaker is: 2mm, _ 2(1000 kg/m3)(0.39 kg) = __ _ m = 27861: 3 27m — pwV [2(039kg) — (1000 kg/m3)(5 x 10-4 1113)] ml P6 4. Let .41 and .42 be the area in the right and left sections of the pipe. a) Durin a. 10 min eriod, the volume of water delivered into the atmos here is g P P equal to: v = Amt = 1r(d1/2)2v1t = 1r(0.03 111/2)2(15m/s)(10 x 605) = 6.36 m3 (b) By the equation of continuity, mAl = 112.42. Hence. the ﬂow speed of the water in the left section is given by: _ .41 _ d1 2 _ 0.03m ’ _ v2 — t1 — 11 — (15m/s) 0.05111 .— 5.4 m/s (0) The Bernoulli' equation states that: 1 1 P, + 5,»)? + Pghl = P2 + 5% + pgh2 Now we have h; = ’12 and P1 = P0 =at1nospheric pressure. So the pressure in the left section of the pipe is: 1 P2 = P0 + 5/10}? - 11%) l = 1.01 x 105 Pa+ 5(1000kg/n13)[(15m/s)2— (5.4111/s)2] : 1.99 x 105 Pa 5. (a) The Bernoulli' equation states that: 1 1 P1 + ﬁlm? + Pghl = P2 + 5W3 + We Now that we have vl = Om/s and P1 = P2 =atmospheric pressure. Thus we obtain the speed of the stream when it leaves the pipe: ‘02 = V 29011 -’12)= V2gh 2 which is directed along the horizontal direction. Suppose it takes time t for the water to reach the ground after it leaves the pipe. Then we have: {1-)}:th = t: m 2 9 Of course. there would be no change in the horizontal speed of the water after it leaVes the pipe. So the distance from the base of the tank to the point at which the resulting stream strikes the ﬂoor is equal to: 1- = v2: = «291: m = 2\/h(H — h) 9 (b) From the expression 1‘- = 2‘/h(H — h), we realize that (H — h) below the surface of water is another possible location of the hole. (c) To find the maximum distance 1m that the emerging stream strikes the ground, we should consider the. function: 1’ = 4h(H — h) = —4h2 +4Hh It is a quadratic function with the form y = ah2 + bh + c. For such function. the maximum value occurs when 12 = —b/(2a). So the emerging stream strikes the ground at the maximum distance 13m“ when b 4H H h. : __ = _ ’E= 2(—4) 2 6. (a) We assume that the weight of the platinum ring is negligible. Then the surface tension is the force to lift the ring. divided by the length of surface that is being pulled. Surface tension will act at both edges of the ring. Thus I_ F _F “Emma—H (b) If F = 8.40 X 10‘3N and r = 2.8 cm at 30°C, then F 8.40 x 10‘3 N : — : — = 2.39 10-2N 7 41rr 47r(‘2.8 X 10‘2 m) x /m ...
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