1. The problem statement, all variables and given/known data Water flows at 0.25 L/s through a 9.0-m-long garden hose 2.0 cm in diameter that is lying flat on the ground. The temperature of the water is 20 ∘C. What is the gauge pressure of the water where it enters the hose? Side question: does the velocity of the water flow need to be converted to m/s rather than L/s? Upon conversion this becomes m^3/s, and I am not sure if this is suitable units for the velocity. R = .01m n = 1.0*10^-3 Pa*s L = 9m v= 0.25L/s 2. Relevant equations Q = (pi*R^4*p)/8*n*L (p = pressure change; n = coefficient of viscosity, in this case 1.0*10^-3 due to the temperature of the water; L= length of the tube, 9m; R = .01m) Q = v(avg)*A (A = cross sectional area, of a tube is pi*r^2) v(avg) = (R^2/8nL)/p. 3. The attempt at a solution I am unsure of where to begin. The first equation I would think to use is v(avg) = (R^2/8nL)*p so that I could plug this into Q = v(avg)A but the first equation requires p and that is what I am solving the problem for. To start, would it be suitable to use the velocity given and change the units to m^3/s? If I take this route, I get: v = 0.25L/s 1000L = 1m^3 0.25L/s * 1m^3/1000L = 0.00025 m^3/s = v Q = v*A; A = pi*R^2 A = pi* 0.01^2; A = 0.000314 m^2 Q = 0.00025m^3/s * 0.000314 m^2; Q = 7.85 * 10^-8 Q = (pi*R^4*p)/8nL p= [(8nL)*Q]/pi*R^4 p = [(8 * (1.0*10^-3 Pa*s)) * 9m]/pi * 0.01^4 p = (0.072)/3.14*10^-8 p = 2292993 or 2.293 * 10 ^6. This answer is incorrect and I am weary of the steps I took to get there. Can someone please point me in the correct direction? The first concern I have is converting L/s for velocity. Should I be leaving velocity in terms of L/s or changing them into m^3/s as I have done here? Also, is this velocity able to be used at all? The formula calls for average velocity, but with the formula for average velocity, I am unable to solve because it calls for the use of p, which I do not have. Any help is appreciated. Thank you in advance.