Which graphic most closely depicts reality?

1. Physics 1710—Warm-up Quiz Answer Now ! 10 39% 55 of 140 0 Which graphic most closely depicts reality? • None of the above. A B C

2. Physics 1710—Chapter 14 Fluid Dynamics 0 Hydrostatic Column A B C

3. Physics 1710—Chapter 14 Fluid Dynamics 0 Review: Pressure is the force per unit area. P =F/A Unit of pressure [Pacal] = [N]/[m2] The hydrostatic pressure is P = Po + ρgh Vx∝ P =Po + ρgh

4. Physics 1710—Chapter 14 Fluid Dynamics 5 kg No Talking! Confer! Think! 0 ρ = 5000 kg/m3 What will the scales read? 20 kg Peer Instruction Time

5. Physics 1710—Chapter 14 Fluid Dynamics 0 41% 58 of 140 Answer Now ! 0 What will the scales read? • Top 4 kg, bottom 20 kg • Top 5 kg, bottom 20 kg • Top 4 kg, bottom 21 kg • Top 0 kg, bottom 25 kg • None of the above

6. Physics 1710—Chapter 14 Fluid Dynamics mg-ρfluid g(m/ρ) ρfluid g(m/ρ) - Mg - mg - ρfluid g(m/ρ) -[Mg+ρfluid g(m/ρ)] 0 Review: Archimedes’ Principle: Fbuoyant = ρfluid g V

7. Physics 1710—Chapter 14 Fluid Dynamics P P 0 F = AΔP - mg= A ρfluid g Δh – mg F = Fbouyant - mg Fbouyant = ρfluid g V Archimedes Pressure A V - mg

8. Physics 1710—Chapter 14 Fluid Dynamics P P 0 Fbouyant = ρfluid g V Archimedes Pressure V - ρfluid V g

9. Physics 1710—Chapter 14 Fluid Dynamics 0 Archimedes Pressure Archimedes’ Principle: Fbuoyant = ρfluid g V The buoyant force of a submerged body is equal to the weight of the fluid displaced.

10. Physics 1710—Chapter 14 Fluid Dynamics 0 Why does the atmosphere extend above P = Po + ρghh = 101 kPa/(1.24 kg/m3x 9.8 N/kg) h = 8.3 km ~ 5 miles?

11. Physics 1710—Chapter 14 Fluid Dynamics No Talking! Confer! Think! 0 Why does the atmosphere extend above 8.3 km? Peer Instruction Time

12. Physics 1710—Chapter 14 Fluid Dynamics 0 Why does the atmosphere extend above 8.3 km? dP = - ρg dy But ρ = ρo (P/Po) So P -1 (dP/dy) = - g ρo /Po P = Po exp[ - (gρo/ Po)y] At y = 8.3 km, P = 0.37 Po ~ 37 kPa

13. Physics 1710—Chapter 14 Fluid Dynamics 0 V constant if fluid is incompressible. A1v1 = A2v2 Continuity

14. Physics 1710—Chapter 14 Fluid Dynamics 0 ΔW = ½ m v2 + mg y dW/dV = ½ ρ v2 +ρg y = d (Fy)/dV = dF/dA =ΔP ½ ρ v2 +ρg y =ΔP Work done on fluid

15. Physics 1710—Chapter 14 Fluid Dynamics 0 Bernoulli’s Equation: P + ½ ρv2 + ρgy = constant = total energy per unit volume

16. Physics 1710—Chapter 14 Fluid Dynamics 0 Summary: Pressure is the force per unit area. P =F/A Unit of pressure [Pacal] = [N]/[m2] The hydrostatic pressure is P = Po + ρgh Archimedes’ Principle: Fbouyant = ρfluid g V Equation of Continuity: A1v1 = A2v2 Bernoulli’s Equation:P + ½ ρv2 + ρgy = constant.

17. Physics 1710—Chapter 15 Oscillatory 0 • F = m a • - PA = ALρ d2y/dt2 • gρy = L ρd2y/dt2 • d2y/dt2 = --(g/L) y • y = Yo sin(ω t);ω = 2π√ (g/L) y What happens in the following case: ρ L

18. Physics 1710—Chapter 15 Oscillatory Oscillatory Motion: y = Yo sin(ω t); sinusoidalω = 2π√ (g/L); angular frequency Fy = - ky Fy = md2y/dt2 Then y = Yo sin(ω t) with ω = 2π√ (k/m)

19. Physics 1710—Chapter 14 Fluid Dynamics 10 0% 0 of 1 Answer Now ! Where should the fulcrum be place to balance the teeter-totter?

20. Physics 1710—Chapter 14 Fluid Dynamics 10 0% 0 of 1 Answer Now ! Which way will the torque ladder move? • Clockwise • Counterclockwise • Will stay balanced