ou are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You pl
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Answer:
The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²
The shear stress at a distance 0.5-in away from the pipe wall is 0
Explanation:
Given;
pressure drop per unit length of pipe = 0.6 psi/ft
length of the pipe = 12 feet
diameter of the pipe = 1 -in
Pressure drop per unit length in a circular pipe is given as;
make shear stress (τ) the subject of the formula
Where;
τ is the shear stress on the pipe wall.
ΔP is the pressure drop
L is the length of the pipe
r is the distance from the pipe wall
Part (a) shear stress at a distance of 0.3-in away from the pipe wall
Radius of the pipe = 0.5 -in
r = 0.5 - 0.3 = 0.2-in = 0.0167 ft
ΔP = 0.6 psi/ft
ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²
Part (b) shear stress at a distance of 0.5-in away from the pipe wall
r = 0.5 - 0.5 = 0
Answer:
-1.2 kg - m/s
Explanation:
And we need to find out the change in momentum of the body . Here ,
- velocity before collision (u) = 10m/s
- velocity after collision (v) = 2m/s .
We know that momentum is defined as amount of motion contained in a body . Mathematically ,
Therefore change in momentum will be,
Since the direction of velocity changes after the collision , the velocity will be -2m/s .