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Ostrovityanka [42]
3 years ago
6

Which sentence is an example of formal language?

Engineering
1 answer:
mote1985 [20]3 years ago
6 0
D sounds more formal than the rest.
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A machine raises 20kg of water through a height of 50m in 10secs. What is the power of the machine.​
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hhahhhwghwhwhwhwjwnwjnnnnwnwwnw

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jwkwkkwoiwiwiwiwiwowwiwowowiiiiwuuwuwgeevehehsvhsvwhbhhehehwgjjwhwhjwjqwjjuuuwi####!\\\\e

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The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome
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(a) Relative Humidity = 48%,

Specific humidity = 0.0095

(b) Enthalpy = 65 KJ/Kg of dry sir

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(c/d) 12.78 degree C

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For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's
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The U.S. Coast Guard requires certain equipment be carried onboard a vessel based on the vessel's length. How must an operator m
nevsk [136]

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3 0
3 years ago
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0
3 years ago
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