Week 2 Notes
Week 2 Notes PHYS 5B
Popular in Intro to Physics II
Popular in Physics 2
This 8 page Class Notes was uploaded by Shanee Dinay on Monday February 8, 2016. The Class Notes belongs to PHYS 5B at University of California - Santa Cruz taught by A.Steinacker in Fall 2015. Since its upload, it has received 20 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.
Reviews for Week 2 Notes
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/08/16
Day 4 1/11/2016 Section start this week Her office hours Monday 57:00 Office Hours in ISB 235 Reading 13.5, 13.6, 13.7 Ex. Measured weight of ball In water: Wa 6.5 Water poured out + beaker = 16.5. Beaker = 4.5 Wf eaker beaker = 12 FB= Wf W W = 18.5 6.5 = 12 0 a 1) Neutral Buoyancy FB= W0→ m0 = V00 FB= Wf mf Vff So Vff 00 object is completely underwateρg = vρg so volumes are equal = v so p = p ff 00 f 0 f 0 object can float if density is the same as fluid 2) density of object is larger than fluid, object sinks ρ0 ρftil0= f W0 FB 3) FB F0 V0 Vf ← when ρf V0 density object is less than fluid, object floats 4) object on the surface: FB= W0 W 0 m0 = V00 Vfis the fluid displaced < object Wf mf Vff V00= Vff when floats Ex. Ice Floating 3 3 Vf ρice 7 kg/m pf= 1030 kg/m salt water V o=? V 3 F = W → V ρ = Vρ f= ρo = 917 kg/ = 0.89 B 0 00 ff V o ρf 1030 kg/m Ex. Is water going to spill? When ice melts, what is volume of water added V V = m V = 0.89V = V = m • = f m ρw f 0 0ρf ρ0 ρf ρf Ex. more buoyant versus less buoyant FB2 FB1 object 1, more buoyant, but buoyant force smaller V0 V0 ρf V f, small volume, ratio has to be large Vf = ρ0 ρfhigher, higher fluid density ρ0lower, lower density Archimedes crown example mc ρfc ρcrown gold changes buoyancy F = BC fc Wcf = ρc fc=w C ρc ρf FBGW fg WGρG FBG FBCρ G> C Two containers: observe object floats ρu= ? FBWW 0 FBu W0buoyant force does not change = ρ Vg = ρVg wfwρ uf = ρu ρw w fu Steal Boat 3 3 ρS= 8000 kg/m pw= 1000 kg/m float = force balance = B= W0 FB= Vff= A•dρw = m g + mg = Adsρ g + (d dsAρg) s a s a Adρw = Adsρs + (dds)Aρa d(ρw a = ds(s a ρs−a 1 − small d = ds w −a= ds pw/ps − smalld = 8ds Ex. Aluminum Sphere m ρ W = W F = W Vρg = W Vρg= W Vρg= W 0fg = W W a 0 B 0 ff 0 0f 0 0f 0 ρ0f 0 0 0 https://ecommons.ucsc.edu/access/content/group/3a4420f5de24449baa633cc45869e478/LE CTURE%20NOTES/Class%204.pdf Day 5 1/13/2016 Physics 5b Wed Section ISP 165 Homework Problem Changed Reading 13.8, 13.9 Fluid Flow Ideal Fluid Model 1) Incompressibility: density is constant for a gas to be incompressiblg<< speed of sound 2) Inviscid (no viscosity) resistance to flow, stickiness 3) flow is steady 4) Irrotational Take a flow tube mass conservation dm = dm 1 2 dm1 dm2 ρdV1= A1x ρA1 1 ρA11t ρA11t = ρA22 A11= A22continuity equation as A goes up, volume goes down Ex. =ρVAcosΘ mdot = ρV•A = mass flux = mass flow rate kg kg m 2 [m] = s m3• s• m Ex. curved surface ∫ ρV •dA = dm area 2 2A11= A222 E1= E2 mv1 mv1 mv2 2 2 E1 2 + mgh 2 + mgh = 2 → v2= v1+ 2gh mv2 A1 1 2 2 E2 2 v2= A v1 A v1= v1+ 2gh 2 2 2gh 2gh v1= (A1/2 ) −1= 1 − 1A2/A ) √ √ Bernoulli's Equation F1= P11 F2= P22 dm W1= F1 1= P11 1= P11= P1 ρ dm W2= F2 2 P22 2= P2 ρ W = W 1+ W2= ΔE = E2 E1 dmv2 dmv1 E2= 2 + dmgh 2 E1= 2 + dmgh 1 v dmv2 dm 2 + dmgh 1 dmgh = P P dm 2 2 2 1 1 ρ2 2 ρ 2 ρv2 ρ1 Static pressure, dynamic P2+ 2 + ρgh2 P1+ 2 + ρgh1 Ex. ` A1goes down → V 1goes up ρV 2 ρV2 P1+ 2 = P2+ 2 Ex PA= PC PB no so → PB= P0+ ρgh > PA PA PB= PA ρgh put a tube above the three tubes, and have it narrow above the B tube vBgoes 2p because A2goes down P + ρovB = P + ρovA B 2 A 2 Day 6 11/15/2016 Phys 5m Fluids Lab Next Week Homework 22 Posted Today C.E: v11= v22 ρv ρv2 BE: 1+ P1= 2 + P2 2 2 Ex. from picture above D1 2cm, D2= 1cm, h = 10cm A1 v11= v22→ v2= v1A 2 2 2 ρv1+ P = ρv1( A1) + P (Eq 1) 2 1 2 A2 2 PA= P1PB= P2 ρgh P1= P2+ ρgh No2 putting 2verything in Eq 1 ρv1 ρv1 A1 2 + P2 2 A2 + P2 A A ρo1 + 2ρgh = ρo1 1) 1o1 A2 A2 2 A1gh = v ρ(( A1) 1) A2 1A2 A2 2(ρoρ )gh v1= (A /A ) −1 √ 1 2 3 3 2 Solv: pw= 1000kg/mp o 1.28kg/m g = 9.8m/s h = 0.1m v = 10.1 m/s 1 Q = A v cross sectional area times flow speed 11 Ex D = 2m D = 4mm = 4 x 10h3 = 1m 1 2 1 v2= ? v1= ? ρv ρv Av = Av P + 1 + ρgh = P + 2 + ρgh 11 22 0 22 1 o 2 2 ρ (A2)v + ρgh = ρv2 + ρgh 2 A1 2 1 2 2 2 A2 2 2ρg(h1 2 = ρv21 A ) 1 v2 = 2g(1 22 ) h h = h, difference in height of big container and height hole 2 1−(1 2A ) 1 2 A 2 1 A 2) ~ 1 so v2= √2g(h 1 h 2 1 2 Q = Av = πD•v = 5.57 x 10 m3/s 2 22 4 2 5 v2= √2gh = 4.43 m/s = 3.34 L/min v1= 1.77 x 10 m/s Ex ρv ρv2 1) P + 1 + gh = P + 1 o 2 2 0 o 2 A) P + ρvA ρgh A 22 A B) P + ρvB + ρgh B 2 B ρvC C) Po+ 2 + ρghC D1>> Ds(diameter of the tube) so v << v Area of tube = constant 1 A continuity equation v = v = v = v ← tube speed ρv A B C s 1 & B P 0= PB+ 2s + ρghB ρv 1 & C P 0= P0+ 2s ρghC v s= √2gh c ρ P0= PB+ 2(2ghc + ρghB PB= P0 ρg(hc+ hB best we can do is make P 0= ρg(hc+ hB max hC hB= 10.3m max Adriane’s Notes https://ecommons.ucsc.edu/access/content/group/3a4420f5de24449baa633cc45869e478/LE CTURE%20NOTES/Class%206.pdf
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'