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
The examples of engineering controls is Biohazard waste containers and Spill clean up kits.
What is engineering controls?
An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.
An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.
Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.
Therefore, the Option C and D is correct.
Answer:
true
Explanation:
bc the right plus left valve
Answer:
public static int average(int j, int k) {
return (int)(( (long)(i) + (long)(j) ) /2 );
}
Explanation:
The above code returns the average of two integer variables
Line 1 of the code declares a method along with 2 variables
Method declared: average of integer data type
Variables: j and k of type integer, respectively
Line 2 calculates the average of the two variables and returns the value of the average.
The first of two integers to average is j
The second of two integers to average is k
The last parameter ensures average using (j+k)/2
