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IRISSAK [1]
3 years ago
12

WIL MARK U AS BRAINLIEST HELP PLZ

Engineering
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

The 1st one:Your natural ability

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Murljashka [212]
A and C is the answer to the question. Be 15 years old & get a permit
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3 years ago
What are the height and width of scissors?
timofeeve [1]
Short ones are 4.5 inches but long ones can be up to 8 inches.
8 0
2 years ago
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue
Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0
3 years ago
C++ question <br> Thank you
Salsk061 [2.6K]
You're welcome lol! !!!!!!!!!!!!!!

:):):):)
5 0
3 years ago
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