Answer:
See explaination and attachment.
Explanation:
Iteration method is a repetitive method applied until the desired result is achieved.
Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below
x = pi(x)
Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).
Similarly for second, thrid and so on. approximation
x2 = pi(x1)
x3 = pi(x2)
x4 = pi(x3)
xn = pi(xn-1).
please go to attachment for the step by step solution.
Answer:
Explanation:
Availability:
It define as the probability of system which perform desired task before showing any failure .
The availability can be define as follows
Or we can say that
Availability can also be express as
Where MTBF is the mean time between two failure.
MTTR is the mean time to repair.
Answer:
Time of concentration, ⇒ 1280 min
Peak runoff rate, ⇒4.185 ff³/s
Explanation:
See detailed explanation
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
Answer:
The FSM uses the states along with the generation at the P output on each of the positive edges of the CLK. The memory stores the previous state in the machine and the decoder generates a P output based on the previous state.
Explanation:
The code is in the image.
