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by: Shanee Dinay

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7

# Week 1 Notes PHYS 5B

Shanee Dinay
UCSC
GPA 3.94

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These are the first 3 classes of Intro to Physics. Topics include density, pressure, and hydrostatics.
COURSE
Intro to Physics II
PROF.
A.Steinacker
TYPE
Class Notes
PAGES
7
WORDS
CONCEPTS
physics 2, pressure, Hydrostatics, Physics
KARMA
Free

## Popular in Physics 2

This 7 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 26 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 1 ­ 1/4/2016    Mondays 5­6:30 Sections with the Professor  Adriane Steinacker  First Homework posted this Friday  Lectures are Webcast Login: phys5b Password: Y0ung2sl1t  Class Materials: rulers, compass, scientific calculator  Yellow is her favorite color  1) Density (mass density) lett rho ρ  ρ =    mass over volume [ρ] = kg  v m3 1 L (Liter) = 100cm​   1m​3 = 1000L  1 g kg converting 1 cm3= m 3  Assume air is made of Nitrogen molecules  N​   atomic mass z = 7 7 protons and 7 electrons  2 A = 14 so 7p & 7e  Mass of proton = 1.67 x 10​ ­27 Mass of neutron = M​   p M​N2​ x 14 x 1.67 x 10​ = 4.676 x 10​ ­26 mas of one Nitrogen molecule  1.28 kg air 25​ N​2​= 4.676 x 1 kg = 2.7 x 10​  in one meter cube of air  2) Pressure  ← water bottle  pressure depends on depth  a) why does the water flow out?  dP 1 dP F =  dx F ~  ρ dx pressure force in a fluid  FN b) How do we define pressure? P =  A normal force over the area  Units of Pressure  [P] = mN2 = 1 Pa Pascal  5​       or 1 standard atmosphere 1atm = 1.013 x 10​  Pa        or 1 bar = 100 kPa  circular surface, you can tilt it and it’ll have the same pressure at the  same point in the water cylinder  F​ = (P + dP)A  A ​ F1​+ F​2​ W = 0 →  (­P + dP)A + dP ­ mg = 0  V•ρ  A•dy  dP = ρgdy  dP = ­ρgdy    Check out Steinacker’s notes as well:   https://ecommons.ucsc.edu/access/content/group/3a4420f5­de24­449b­aa63­3cc45869 e478/LECTURE%20NOTES/Class%201.pdf    Day 2 ­ 1/6/2016    Physics 5b  Adriane Steinacker    Reading 13.3, 13.4  David’s sections Tu 11­12:30, 1­2:30, Thu 1­2:30  dP = ρgdy  P0 1 d 1) ρ = const ∫ dP =− pg dy∫  1 P y P​0​ P = ­ρg(d­y) => P(y) = P​ o​+ ρg(d­y)  P(y) = P​o​+ ρgh  hydrostatic formula  Ocean 36,000 feet  3 P1​= P​0​+ ρgd ρ=1030 kg/m​   P​ = 1.013 x 10​ 5  N  + (1030 kg • 9.8 • 36000 ft •  1 g ) = 1.013 x 10​ + 1.11 x 10​   8 N 1​ m2 m 3 s 3.28 ft m2 m2 8 N 5Pa P1​= 1.11 x 10​  /m2.013 x 10​  = 10atmatm  where does P​ 0​ ρgd? d0​= 10m  Ex  Pb−Pt P t   Mount Whitney height 14,505 feet  P = ?  Temperature=const     dP = ­ρgdy P  =  P0 => ρ = P ρ0 lnP ­ lnP​  ρ ρ0 P0 0 P y dP = ­P ρ0 gdy =>  ∫ dP= ­ ∫ρ0gdy => lnP​ 1|​ = ­ ρ0y  P0 P P P 0 P0​ P0 −P g 0 0 P0y P(y) = e   1450 ft kg m P ­0.35​ y =  3.28 ft ρ0 ​ 1.28 m   g = 9.8 s P0 = e​  = 0.58  Otto van Guericke  1 Hemisphere    Ex:    ds = Rdθ dw = rdφ = Rsinθdφ  dA = ds•dw = R​ 2sinθdφdθ    dF​xy​ dFsinθ  dF​  = dF​sinφ  y​ xy​     dF​  = dFsinθsinφ F​ =∫∫P​  • Rsin​θsinφdθdφ F​  = P​ • πR​  y​ y​    0​ y​ 0 ​ 2 5N Calculate the force: Fy​= P • πR​  R = 25cm P​0​= 1.013 x 10​m2  = 2 x 10 N  Hydrostatic Equilibrium  P1​= P0 P2​= P0​+ ρgd P1​+ P2 P2​P​1  P​ = P​  P​ = P + ρgh P​ = P​ + ρgh  0​ 1 1​ 0​ 2​ 0​     1​= 2​so we are in hydrostatic balance  Cap  FNA P1​= P0​+ ρgd P2​=  A P1​= P2    Day 3 ­ 1/8/2016    Reading 13.5, 13.6, 13.7  MSI Times:  Tues 2 ­ 3:15pm Crown 201  Wed 2 ­ 3:10pm Crown 104  Thurs 4 ­ 5:15 ARC 203  Mon 5 ­ 6:15pm SSII 363  Tues 6 ­ 7:15pm ARC 202  Wed 7 ­ 8:15pm Baskin Engineering 169  Ex.    P​ = atmospheric pressure = P​  A​ 0 PB​= ρgh = 1000 m2• 9.8s• 0.1m  P0​= ρgh  5​N mg m 1.013 x 10​ m2= 100 m3 • 9.s2 • 0​ h0​= 10m ← special height, do not forget  If h > 10m water flows out of the bottle  Ex    P​A​= P0  P​B​= P1​+ ρgh P0​= P1​+ ρgh  P​  = P​ 1.013 x 10​5 N  ­ 1000 kg• 9.8 •0.1m = P​  = 1.0032 x 10​   A​ B m2 m 3 s 1​ m2 Ex    h = 50cm = 0.5m  P​0​= ρgh P​0​= P1​ ρgh  5​N m 4​ N P​A​= PB  1.013 x 10​  m2 = ρ • 9.s2 • 0.5m ρ = 2.07 x 10​ m 2  4​kg Mercury density = 1.36 x 10​  m3 h = 76cm = 760mm  Ex. Pressure Gauge & Gauge Pressure    P​A​= P  P​  = P​ + ρgh P​ = P​  P = ρgh + P​  ← we measure ρgh, it is known  B​ 0​ A​ B 0​ P​  = P ­ P​  = ρgh  gauge​ 0​ Ex    caps closing the two containers  P​1​= P2  M g M g M1 M2 P​1​= A1  + P0 P2​= P0​+ A2 P1​= P2  =>  A = A   1 2 1 2 A M πR2 So,  1 =  1 =>  2  = 50  =  1 the ratio is 1:5  A2 M2 πR2 1250 25 M​1​50kg person M​2​= 1250kg car  In the second situation, we have two people on the left cap so the mass if off balance and the  car moves up:  50kg • g 50kg • g 1250kg • g P1​= P0​+  A +  A P2​= P0​+  A  + ρgh  1 1 2 so 50kg • = ρgh h​ = 50kg=  52kg  A1 1​ A1• ρ πR1• ρ 50kg 2​ 2 h = πR1• ρ h = h1​+ h2  M1​= M​2 h1​R​1​ = h2​R​2​  R1 2​ h1​1​ = h2​2​ρ h2​= (R )​h1  2 Ex Buoyancy    ball weight = 18.5  in water weight = 6  water + jar weight = 16.5  jar weight = 4  18.5 = (16.5 ­ 4) + 6  apparent wgt W​ a​= W ­ F​b Upward force comes from difference in pressure across an object in a liquid  Ex    F​ = PA F​ = PA F​ = F ­ F = PA ­ PA = A(P​ ­ P)  1​ 1​ 2​ 2​ b​ 2​ 1​ 1​ 2​ 1​ 2​ F​ = PA + ρgh​A​ ­ PA ­ ρghA  b​ 0​ 2​1​ 0​ 1 1 1​ Fb​= fluid2​­ 1​ = ρf​V Fb​= f​V = weighf​  V = volume of fluid displaced ρg = m

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