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by: Shanee Dinay

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# Week 2 Notes PHYS 5B

Shanee Dinay
UCSC
GPA 3.94

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Week 2 notes of intro to physics 2. Topics include buoyancy and fluids.
COURSE
Intro to Physics II
PROF.
A.Steinacker
TYPE
Class Notes
PAGES
8
WORDS
CONCEPTS
Fluids, Physics, physics 2, Buoyancy
KARMA
25 ?

## 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.

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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: W​a​ 6.5 Water poured out + beaker = 16.5.   Beaker = 4.5  W​f ​eaker ­ beaker = 12 FB​= W​f W​ ­ W​ = 18.5 ­ 6.5 = 12  0​ a​ 1) Neutral Buoyancy    FB​= W0​→ m0​ = V0​0​  FB​= Wf​ mf​ ​Vf​f​ So Vf​f​ ​0​0​  object is completely underwateρg = v​ρg so volumes are equal  = v so p = p  f​f​ 0​0​ 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​ ρf​til0​= f​ W​0​ FB  3) FB​ F0 V​0​ Vf ← when ρ​f​ V0 density object is less than fluid, object floats     4) object on the surface:    FB​= W​0  W​ 0​ m​0​ = V0​0​  Vf​is the fluid displaced < object W​f​ m​f​ ​V​f​f​ V0​0​= V​f​f 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​ = V​ρ​ f= ρo =  917 kg/ = 0.89  B​ 0​ 0​0​ f​f 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​ F​B1 object 1, more buoyant, but buoyant force smaller  V0 V0 ρf V f, small volume, ratio has to be large Vf = ρ0   ρf​higher, higher fluid density ρ0​lower, lower density  Archimedes crown example  mc ρfc ρcrown​​ gold changes buoyancy F​ = BC​ fc​ W​c​f​ = ρc fc=w​ C ρc ρf F​BG​W​ fg​ W​GρG FBG​ F​BC​ρ​ G​> C  Two containers: observe object floats    ρu​= ? F​BW​W​ 0​ FBu​ W​0buoyant force does not change  = ρ​ V​g = ρ​V​g  w​fw​ρ u​f​ = ρ​u​ ρw  w fu Steal Boat    3 3 ρS​= 8000 kg/m​ pw​= 1000 kg/m​   float = force balance = B​= W​0 F​B​= Vf​f​= A•dρw  = m​ g + m​g = Adsρ​ g + (d ­ dsAρg)  s​ a​ s​ a​ Adρ​w​ = Adsρ​s​ + (d­ds)Aρa​ d(ρ​w​­ 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​  ­ Vρ​g= W​  ­ Vρ​g= W​  ­  0fg = W​ ­ W​    a​ 0​ B​ 0​ f​f​ 0​ 0​f​ ​ 0​ 0​f​ ​ 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 dm​1​ dm​2 ρdV1​= A1​x ρA​1​ 1​ ρA1​1​t ρA1​1​t = ρA2​2​  A1​1​= A2​2continuity 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 2A1​1​= A2​22 E1​= E2  mv1 mv1 mv2 2​ 2​ E​1​ 2  + mgh  2 + mgh =  2  → v​2​= v1​+ 2gh  mv2 A1 1 2​ 2​ E​2​  2 v2​= A v1​  A v​1​= v1​+ 2gh  2 2 2gh 2gh v1​=  (A1/2 ) −1​=  1 − 1A2/A ) √ √ Bernoulli's Equation  F1​= P1​1 F2​= P2​2  dm W​1​= F1​ 1​= P1​1​ 1​= P1​​1​= P1 ρ dm W​2​= ­F2​ 2​ ­P2​2​ 2​= ­P2 ρ W = W​ 1​+ W​2​= ΔE = E​2​­ 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 P​2​+  2  + ρgh​2​ P​1​+  2  + ρgh1  Ex.   `   A1​goes down → V​ 1​goes up  ρV 2 ρV2 P1​+  2  = P2​+  2   Ex  P​A​= PC ​ PB no so → P​B​= P0​+ ρgh > P​A  P​A PB​= PA​­ ρgh  put a tube above the three tubes, and have it narrow above the B tube  vB​goes 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: v1​1​= v2​2  ρv ρv2 BE: 1+ P1​=  2 + P2  2 2 Ex. from picture above  D1 ​ 2cm, D​2​= 1cm, h = 10cm  A1 v1​1​= v2​2​→ v​2​= v1A  2 2 2 ρv1+ P​ = ρv1( A1) + P​   (Eq 1)  2 1​ 2 A2 2  PA​= P​1​PB​= P​2​ ρgh P1​= P2​+ ρgh  No2 putting 2verything in Eq 1   ρv1 ρv1 A1 2 + P​2 ​ 2 A2 + P2  A A ρo​1​ + 2ρgh = ρ​o​1​ 1) ­  1o​1​  A2 A2 2 A1gh = v​ ρ(( A1)​ ­ 1)  A2 1​A2 A2 2(ρoρ )gh v1​=  (A /A ) −1 √ 1 2 3 3 2 Solv: pw​= 1000kg/m​p​ 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  1​1​ Ex    D​ = 2m D​  = 4mm = 4 x 10​h​­3  = 1m  1​ 2​ 1​ v2​= ? v​1​= ?  ρv ρv A​v​ = A​v​ P​  +  1 + ρgh​ = P​ +  2 + ρgh​   1​1​ 2​2 0​ 22 1​ o​ 2 2 ρ (A2)​v​ + ρgh​  = ρv2 + ρgh​  2 A1 2​ 1​ 2 2 2​ A2 2 2ρg(h​1​­ 2​ = ρv2​1 ­ A )​  1 v​2 = 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​ = A​v​ =  πD•v  = 5.57 x 10​­ m​3/s  2​ 2​2​ 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  + ρgh​C  D1​>> D​s​(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 + ρgh​B  ρv 1 & C P​ 0​= P0​+ 2s ­ ρghC v​ s​= √2gh c  ρ P0​= PB​+ 2(2gh​c​ + ρghB  PB​= P0​­ ρg(hc​+ hB​  best we can do is make P​ 0​= ρg(hc​+ hB​ max  hC ​h​B​= 10.3m max    Adriane’s Notes  https://ecommons.ucsc.edu/access/content/group/3a4420f5­de24­449b­aa63­3cc45869e478/LE CTURE%20NOTES/Class%206.pdf

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