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fiasKO [112]
3 years ago
14

How many power station do we have​

Engineering
1 answer:
loris [4]3 years ago
5 0

Answer: 9,719

Explanation:

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This is due in a few hours but I appreciate it if it was answered soon so I can rest.
Anna35 [415]
Yessssssssss okkkkkk
8 0
3 years ago
The viscosity of the water was 2.3×10^−5lb⋅⋅s/ft^2 and the water density was 1.94 slugs/ft^3. Estimate the drag on an 88-ft diam
Lina20 [59]

Answer:

hello your question is incomplete attached below is the complete question

answer : Drag force = 1.3 Ib

Explanation:

we have to represent the dimensions of the drag force in terms of FLT

i.e : D = f( <em>d,v,p,u </em>) represented in terms of FLT

D = F , V = LT^-1,  d = L, p = FL^-4 T^2

u = FL^-2 T,  Number of independent terms = 5

attached below is the detailed solution

6 0
4 years ago
A mixture of two gases has a total mass of 80 kg. One gas has a specific volume of 0.8 m^3/kg and occupies a volume of 40 m^3. T
Nady [450]

Answer:

specific volume = 1.025 m³/kg

Explanation:

given data

total mass m1 +m2 = 80 kg

specific volume = 0.8 m³/kg

occupies volume v1 = 40 m³

other gas specific volume  = 1.4 m³

to find out

How much volume is occupied by the second gas and what is the specific volume of the mixture

solution

we know that density is reciprocal of specific volume and here gas is not interacting

so total specific volume is assume so ratio is total volume to total mass

and

specific volume = \frac{1}{\rho}

here ρ is density

so ρ1  =   \frac{1}{0.8} = 1.25 kg/m³

and ρ2  =   \frac{1}{1.4} = 0.714 kg/m³

and

so m1 = ρ1v1 = 1.25 × 40 = 50 kg

and m2 = 80 - 50 = 30 kg

so

v2 = \frac{m2}{\rho 2}

v2 =  \frac{30}{0.714} = 42 m³

so volume occupied by second das = 42 m³

and

specific volume of mixture will be

specific volume of mixture = \frac{v1+v2}{m1+m2}

specific volume = \frac{42+40}{50+30}

specific volume = 1.025 m³/kg

3 0
3 years ago
Does any one like Tom Mcdonld
Arada [10]

Answer:

ye

Explanation:

3 0
3 years ago
The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of
weeeeeb [17]

Answer:

P_{pump}= 68600Pa

Explanation:

We understand that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

In this way,

P_{tank} + P_{pump} = P_{nozzle}

For the pressure we also know that it is given by

P=\rho gh

\rho gh_1 + P_{pump} = \rho gh_2

So,

P_{pump} = \rho g(h2 - h1)

P_{pump}= 1000*9.8(27-20)

P_{pump}= 68600Pa

7 0
3 years ago
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