A liquid of density 1277 kg/m3 flows with speed 2.2 m/s into a pipe of diameter 0.15 m . The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 5.65 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.3 atm . What is the velocity v2 of the liquid flowing out of the exit end of the pipe? Assume the viscosity of the fluid is negligible and the fluid is incompressible. The acceleration of gravity is 9.8 m/s2 and Patm = 1.013

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west