Solutions to Week 9 Assignment

1. f = P_netA = (4.5atm)(1.013 x 10⁵Pa/atm)p(0.010m)² = 140N.
2. The work done by the small piston must equal that done on the car.
        W_p = W_c
     F_pd_p =F_cd_c
         d_p = (F_c/F_p)d_c
              = (A/a)d_c              (F_c/F_p =  = A/a since DP = F_c/A = F_p/a)
              = (100.0)(0.010m) = 1.0m.
3. d = DP/rg = [1000atm/(1025kg/m³)(9.8m/s²)](1.013 x 10⁵Pa/atm) = 10km.
4. P = P_aorta + rgd = 104mm Hg + (1060kg/m³)(9.8m/s²)(1.37m)(760mm Hg/1.013 x 10⁵Pa) = 210mm Hg.
5. The weight of the alcohol displaced is equal to the buoyant force.
   W_alcohol = mg = 1.03N - 0.730N = 0.3N.
   S.G. = r_alcohol/r_w = W_alcohol/r_wV_alcoholg = (0.30N)/[(1.00 x 10³kg/m³)(3.90 x 10^-5/m³)(9.8m/s²)] = 0.78.
6. (a) According to Newton's second law, the weight of the plane is equal to the force due to the pressure difference.
        W = F =DPA = (500Pa)(200m²) = 1 x10⁵N.
   (b) Use Bernoulli's equation with y₁ = y₂
        1/2rv₁² - 1/2rv₂² = DP
                      v₁² - v₂² = 2DP/r
                                v₁ = Ö2DP/r + v₂²
                                     = Ö[2(500Pa)/1.3kg/m³) +(80.5m/s)²
                                     = 85m/s.
7. (DV/Dt_total)/(DV/Dt_original) = (pDP/8Lh)[(D/4)⁴ + (D/4)⁴]/(pDP/8Lh)(D/4)⁴ = (1/256 + 1/256)/(1/16) = 1/8 the original flow rate.
8. Use Stoke's law and Newton's second law.
   SF_y = F_B - F_D - W = 0, so
   m_og - 6phv_t - m_ag = 0
                      6phrv_t  = (m_o - m_a)g
                             v_t  = (m_o - m_a)g/6phr
                                  = 4/3pr³g(r_o - ra)/6phr
                                  = 2rg(0.85r_w - r_a)/9h
                                  = [2(0.0010m)²(9.8m/s²)(0.85 x 10kg/m³ - 1.20kg/m³)]/[9(0.12Pa·s)]
                                  = 1.5cm/s.
9. (a) F =DPA = (2g/r)pr² = 2p(0.070N/m)(0.02 x10^-3m) = 9 x 10^-6N.
   (b) Use Newton's second law.
        SF = 6F - mg = 0
        m = 6F/g = 12pgr/g = 12p(0.07N/m)(0.02 x 10^-3m)/(9.8m/s²) = 5mg.