A) In the case of the Boundary Thickness Layer we use the given formula,
We know as well that,
Re = Número de Reynolds = 
Where,
U = velocity
= kinematic viscosity
For water, kinematic viscosity, 
So, 
B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.
Thus,
C. Wall shear stress,
For water, dynamic viscosity, 
= 2.344*10^-5 lbf-s/ft^2
Answer:
4.37 * 10^-4 J
Explanation:
Energy stored :
mgΔl / 2
m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m
Δl = mgl / πr²Y
Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m
Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10
Δl = 98 / 3.5 * π * 10^6
Δl = 0.00000891267
Energy stored :
mgΔl / 2
(10 * 9.8 * 0.00000891267) / 2
= 0.00043672083 J
4.37 * 10^-4 J
Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:
where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,
<u>F = 51.3°</u>
