Which element family usually shown separated from the main part of the periodic table
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
Answer:
a) 6.4 kJ
b) 43.4 kJ
Explanation:
a)
= Heat absorbed = 37 kJ
= Coefficient of performance = 5.8
= Work done
Heat absorbed is given as
=
37 = (5.8)
= 6.4 kJ
b)
= work per cycle required
=
+
= 37 + 6.4
= 43.4 kJ
Answer:
The horizontal distance of Sosa is 276.526 ft or 84.28 meter.
Explanation:
As shown in the figure, let point O is the starting point of Sosa. She travels 361 ft at an angle 50 degree with the horizontal.
sin 50 =
0.7660 = h / 361
h = 276.526 ft
h = 84.28 meter
The horizontal distance of Sosa is 276.526 ft or 84.28 meter.
