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stealth61 [152]
3 years ago
8

3. Air at 1 atm and 20 0 C flows tangentially on both sides of a smooth flat plate of width W=10 ft and length L=10 ft in the di

rection of flow (x-direction). The velocity outside the boundary layer is constant at 20 ft/s. a) What is the laminar boundary layer thickness at x=1 ft? b) The boundary layer varies as x to what power? c) If the plate is 100 ft long, does the boundary layer thickness change at 1 ft?
Engineering
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

ExplanationAir at 1 atm and 20°C flows tangentially on both sides of a thin, smooth flat plate of ... ... Smooth Flat Plate Of Width W = 10 Ft, And Of Length L 3 Ft In The Direction Of The Flow. The Velocity Outside The Boundary Layer Is Constant At 20 Ft/s.

:

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3 years ago
A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is
jonny [76]

Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}

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The time required to travel is:

t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }

t = 277.477\,s\,(4.625\min)

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

7 0
3 years ago
The aluminum rod AB (G 5 27 GPa) is bonded to the brass rod BD (G 5 39 GPa). Knowing that portion CD of the brass rod is hollow
Temka [501]

Answer:

Qcd=0.01507rad

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The full details of the procedure and answer is attached.

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3 years ago
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nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

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2 years ago
The B-pillar may also be called the:
slega [8]

Answer:

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