14. Earth goes around the sun at an astonishing speed of about -hold your breath-107 460 643 meters per second. *
1 answer:
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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
Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
(1)
where
a is the deleceration
S is the distance covered after a time t
is the velocity at time t
is the initial speed of the car
The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and
(the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve the equation (1) to find a:
And the negative sign means it is a deceleration.
where
a is the deleceration
S is the distance covered after a time t
The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and
And the negative sign means it is a deceleration.