( ) 2 ( ) 9.80 m s Pa kg m 9.80 m s m

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1 Cpter luid Mecnics P. M = ρ V = ( k π( iron M = 0. k P P = = = 6. 0 N π ( P. Te Ert s surfce re is Tis fce is te eit of te ir: so te ss of te ir is π R. Te fce pusin inrd over tis re ounts to 0 0 ( π = P = P R ( π = = P R 0 ( π R ( N π ( P0 8 = = = k 9.80 s 5 P.6 ( P= P + ρ = P 0 k 9.80 s P =.0 0 P Te ue pressure is te difference in pressure beteen te ter outside nd te ir inside te subrine, ic e suppose is t.00 tospere. 7 Pue = P P0 = ρ =.00 0 P Te resultnt inrd fce on te ptole is ten = P = π = 7 5 ue.00 0 P N P.7 = 80.0 k 9.80 s = 78 N Wen te cup brely suppts te student, te nl fce of te ceilin is zero nd te cup is in equilibriu. ( P = = P = 78 = = = P IG. P.7

2 P. Te pressure on te botto due to te ter is So, Pb b = ρ z= b.96 0 P = P = On ec end, Pvere ( On te side, Pvere ( N don = = P 0.0 = 96 kn outrd = = P 60.0 = 588 kn outrd P.6 ( Usin te definition of density, e ve 00 ter = = = ρter 5.00 c.00 c 0.0 c Sketc t te rit represents te sitution fter of ercury te ter is dded. volue s been displced by ter in te rit tube. Te dditionl volue of ercury no in te left tube is. Since te totl volue of ercury s not cned, = = IG. P.6 ( t te level of te ercury ter interfce in te rit tube, e y rite te bsolute pressure s: P= P0 + ρter Te pressure t tis se level in te left tube is iven by P= P0 + ρh( + = P0 + ρter ic, usin eqution ( bove, reduces to ρh + = ρter ρter = ρ + / H Tus, te level of ercury s risen distnce of = (.00 c ( 0.0 c (.6 c ( + 0.0/ 50.0 P.7 P0 ρ.66 0 P : = 0.90 c bove te iinl level. P= P0 +Δ P0 = P = P Δ = Δ = 5 5 P.0 ( Te blloon is nerly in equilibriu: = B = 0 y y eliu pylod ρir V ρeliu V pylod = 0 Tis reduces to ( ρ ρ V.9 k 0.79 k ( 00 pylod ir eliu pylod Siilrly, = = = k

3 ( ρ ρ V.9 k k ( 00 pylod ir ydroen pylod = = = 80 k Te surroundin ir does te liftin, nerly te se f te to blloons. P.7 ( ccdin to rciedes, B= ρtervter =.00 c But B Weit of block ρ V ( c ( 0.0 c = = = = ood ood = ( = so = 0.0( = 7.00 c B= + M ere M = ss of led.00 ( 0.0 = 0.650( M M = ( ( 0.0 = 0.50( 0.0 = 800 =.80 k P.7 lo rte Q = s = v v Q / s π (0.0 = = =.6 s P.9 In te reservoir, te ue pressure is.00 N Δ P = = ro te eqution of continuity: v = v ( 5 ( v =.00 0 v v = (.00 0 Tus, is neliible in coprison to. v Ten, fro Bernoulli s eqution: v v P P P + ρv + ρy = ρv + ρy P.5 Wen te blloon coes into equilibriu, e ust ve B y =, blloon, He, strin = 0 He, strin is te eit of te strin bove te round, nd B is te buoynt fce. No, blloon blloon, He He ir = = ρ V B= ρ V nd, strin strin = L IG. P.5

4 Terefe, e ve ρ L irvblloon ρh ev strin = 0 ivin = ( ρ ρ V ir H e blloon strin ( π L k 0.00 / 0.50 k = (.00 = k P.5 Consider te dir nd pply Bernoulli s eqution to points nd B, tkin y = 0 t te level of point B, nd reconizin tt v is pproxitely zero. Tis ives: P + ρ( 0 + ρ( Lsinθ L B Vlve = PB + ρvb + ρ( 0 θ No, reconize tt P = PB = Ptospere since bot points re open to te tospere (nelectin vrition of tosperic pressure it ltitude. Tus, e obtin IG. P.5 vb = ( Lsin θ = ( 9.80 s 0.0 (.00 sin 0.0 vb =. s No te proble reduces to one of projectile otion it vyi = vb sin 0.0 = 6.6 s. Ten, v = v + ( Δy ives t te top of te rc (ere y = yx nd v = 0 yf yi 0 = ( 6.6 s + ( 9.80 s ( yx 0 y =.5 ( bove te level ere te ter eeres. x yf P.55 t equilibriu, y = 0 : B sprin, He, blloon = 0 ivin kl B ( = = + sprin He blloon But B= eit of displced ir = ρ V ir nd He = ρhe Terefe, e ve: kl = ρirvρhev blloon V IG. P.55 L = ( ρ ρ V ir H e blloon k

5 ro te dt iven, L =.9 k 0.80 k k 9.80 s 90.0 N Tus, tis ives L = 0.60 P.66 Let s stnd f te ede of te cube, f te dept of iersion, ρ stnd f te density of te, ρ stnd f density of ter, nd ρ stnd f density of te lcool. ( ccdin to rciedes s principle, t equilibriu e ve ρ s = ρ s = s ρ ρ Wit ρ = k ρ =.00 0 k nd s = 0.0 e et = 0.0( 0.97 = We ssue tt te top of te cube is still bove te lcool surfce. Lettin stnd f te tickness of te lcool lyer, e ve ρ s + ρ s = ρ s so ρ ρ = s ρ ρ Wit nd ρ = k = 5.00 = =.. e obtin (c Here = s, so rciedes s principle ives ( ( ( ρ ρ ( ρ ρ ( ( ρ s + ρ s s = ρ s ρ + ρ s = ρ s = s = 0.0 =

6 P.67 Enery f te fluid-ert syste is conserved. ( K+ U +Δ E = K+ U i ec f L = + v 0 v = L = s =. s P.68 ( Te flo rte, v, s iven y be expressed s follos: 5.0 liters = = 0.0 s 0.8 liters s 8 c s Te re of te fucet tp is π c, so e cn find te velocity s 8 c s flo rte v = = = 65 c s =.65 s π c We coose point to be in te entrnce pipe nd point to be t te fucet tp. v = v ives v = 0.95 s. Bernoulli s eqution is: nd ives P P = ρ v v + y y ρ P P = 0 k.65 s 0.95 s 0 k 9.80 s.00 + ( ( ( P P P ue = =. 0 P P.7 ( diverin stre lines tt pss just bove nd just belo te ydrofoil e ve P + ρy + ρv = P + ρy + ρv b t t t b b Inin te buoynt fce ens tkin yt yb Pt + ρ( nvb = Pb+ ρv b Pb Pt = ρvb( n P P v n Te lift fce is ( b t = ρ b(

7 liftoff, ρvb ( n = M v b M = ρ ( n Te speed of te bot reltive to te se ust be nerly equl to tis speed of te ter belo te ydrofoil reltive to te bot. (c v ( n ρ = M ( 800 k 9.8 s = = 9.5 s k.70

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