The wind tunnel speed that must be used to test the car 1 / 5 model of a car that normally runs at 30 mph is 150 mph
Re = u L / ν
Re = Reynold's number
u = Flow speed
L = Characteristic linear dimension
ν = Kinematic viscosity
ν
= 1.516 * 
m² / s
ν = ν
= 1.516 * m² / s
L = L / 5
u = 30 mph
Re = u L / 5 * 1.516 *
Re = 30 L / 1.516 *
In tunnel,
Re = Re
u L / 5 * 1.516 * = 30 L / 5 * 1.516 *
u = 30 * 5
u = 150 mph
Therefore, the wind tunnel speed that must be used to test the car is 150 mph
To know more about Reynold's number
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What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>
<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>
The answer comes to: <span>−1.36×10^−19 J</span>
You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
thw answer is C. 2
i do not know why! Sorry!
i just took the test and it was right
