Answer:
b. 1232.08 km/hr
c. 1.02 kn
Explanation:
a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.
See attachment for the remaining.
Answer:
I would say do it at an even pace
Explanation:
Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast
Answer:
thickness1=1.4m
thickness2=2.2m
convection coefficient=0.33W/m^2K
Explanation:
you must use this equation to calculate the thickness:
L=K(T2-T1)/Q
L=thickness
T=temperature
Q=heat
L1=0.04*(0--350)/10=1.4m
L2=0.1(220-0)/10=2.2m
Then use this equation to calculate the convective coefficient
H=Q/(T2-T1)
H=10/(250-220)=0.33W/m^2K
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%