Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:
The boundary layer thickness is equal to:
ft
The shear stress is equal to:
b) If the railing edge is 2 ft, the Reynold number is:
The boundary layer is equal to:
The sear stress is equal to:
c) The drag coefficient is equal to:
The friction drag is equal to:
Answer:
308657
Explanation:
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Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
Answer:
Explanation:
Yes I agree with the statement .
When a person who is perfectly insulated from the earth , touches a Van de Graaff , his body acquires charge . when the hair acquires it, it stands out due to mutual repulsion . It is to be noted here that at pointed areas on a surface , there is larger accumulation of charge. Accumulation of charge is greater at hair tops .
It is also a general observation that when a bird sits on high tension wire , his feather stands out due to the same reason.
what? Is there a picture you can show us with this question
