Answer:
τ = 0.25 lbf/in²
Explanation:
given that the oil viscosity, μ = 2.415 lb/ft-s
gap between plates = 1/4 inches = 1/4*12 = 1/48 ft
recall from newtons law of viscosity;
shear stress τ = μ du/dy =
τ = (2.415 lb/ft-s) (10 ft/s)/(1/48) ft
τ = 1159.2 lb/ft-s²
we know that, 1 slug = 32.174 lb
lb = 1/32.174 slug
∴ τ = 1159.2/32.174 slug/ft-s² = 36 slug/ft-s²
τ = 36 slug/ft-s²
multiply both the numerator and denominator by ft, this gives
τ = 36 slug-ft/ft²-s²
τ = 36 lbf/ft² where 1 slug-ft/s² = 1lbf
since 1 ft = 12 inch = 1 ft² = 12² in² = 144 in²
∴ τ = 36/144 lbf /in² = 0.25 lbf/in²
τ = 0.25 lbf/in²
software engineers
hardware engineers
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biomechanical engineers
Answer:
It will be B
Explanation:
Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.
Answer:
2543 k
Explanation:
This problem can be resolved by applying the first law of thermodynamics
<u>Determine the adiabatic flame temperature</u> when the furnace is operating at a mass air-fuel ratio of 16 for air preheated to 600 K
attached below is a detailed solution
cp = 1200
Answer:
R min = 28.173 ohm
R max = 1.55 × 
ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = 
...........1
so R will be
R = 
....................2
put here value
so for t min 11.5 μs
R = 
R min = 28.173 ohm
and
for t max 6.33 ms
R max = 
R max = 1.55 × ohm
