Answer:
R min = 28.173 ohm
R max = 1.55 ×
ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) =
...........1
so R will be
R =
....................2
put here value
so for t min 11.5 μs
R = 
R min = 28.173 ohm
and
for t max 6.33 ms
R max =
R max = 1.55 ×
ohm
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
<u>Determine the design flow ins ft^3/s </u>
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
Answer:
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<h3>• Define Instance :- </h3>
- The data stored in database at a particular moment of time is called instance of database.
<h3>• Define Schema :- </h3>
- Database schema defines the variable declarations in tables that belong to a particular database
<h3><u>The value of these variables at a moment of time is called the instance of that database.</u></h3><h3 />
Explanation:
<h3>Hope this helps !! </h3>