Please help, I am stumped.
2 answers:
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Answer:
The magnitude of change in momentum is (2mv).
Explanation:
The momentum of an object is given by the product of mass and velocity with which it is moving.
Let the mass of ball is m. A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it struck the wall.
Initial speed of the ball is v and final speed, when it rebounds, is (-v). The change in momentum is given by :
p = final momentum - initial momentum
So, the magnitude of change in momentum is (2mv).
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