An oil with a viscosity of mu = 040 N middot s/m^2 and density rho = 900 kg/m^3

Question

An oil with a viscosity of mu = 040 N middot s/m^2 and density rho = 900 kg/m^3 flows in a horizontal pipe of diameter D = 0.20 m and length, L = 10m. The basic physics behind laminar fluid flow says that the flowrate Q, varies directly as the pressure drop, delta p and the radius, r, raised to the 4^th power and inversely as the viscosity mu, and length of pipe L. The proportionality constant is pi/8. a) Write this as an equation. b) Determine the required pressure drop, delta p in N/m^2 to produce a flowrate of Q = 2.0 times 10^-5 m^3/4. You completed your degree and just took a position earning $60,000 per year. You are determined to maintain your salary at a rate higher than the expected 3% inflation per year. What must your salary be 5 years from now to just maintain your standard of living. A factory machine produces a sound with an intensity of 20 times 10^-3 W/m^2. What would be the decibel rating if 5 of these machines occupied the same room? Does it exceed the dB level of 120 for a rock concert? dB = 10 log_10 (intensity/reference intensity) Use a reference intensity of 1 times 10^-12 W/m^2 which is the threshold of hearing. Solve for x in the equation: x^2 + 13x = 0 Solve for x in the equation: x^2 - 4x + 16 = 4x Solve for x and y in the following system of equations: 3y + 6x = 9 2y + 10x = 4 Solve for (X_1, Y_1)and(X_2, Y_2) for the following system of equations: y = -0.5x + 5 x^2 + y^2 = 100

Explanation / Answer

7) x^2 + 13 x = 0

x(x + 13) = 0

x = 0 or x = -13

8) x^2 -4x +16 = 4x

x^2 -8x + 16 =0

(x-4)^2 = 0

x = 4

9)

3y + 6x = 9 ,

2y + 10 x= 4

solving we get

x = -1/3 , y = 11/3

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