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Llana [10]
2 years ago
8

5 kg of a wet steam has a volume of 2 m3

Engineering
1 answer:
Verizon [17]2 years ago
4 0
Have you seen the state of her body. Mad. If I wear it I ain’t wearing a jolly. ADEOLA wanna ride in the gysa
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Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo c
brilliants [131]

Answer:

The algorithm is as follows:

1. Declare Arr1 and Arr2

2. Get Input for Arr1 and Arr2

3. Initialize count to 0

4. For i in Arr2

4.1 For j in Arr1:

4.1.1 If i > j Then

4.1.1.1 count = count + 1

4.2 End j loop

4.3 Print count

4.4 count = 0

4.5 End i loop

5. End

Explanation:

This declares both arrays

1. Declare Arr1 and Arr2

This gets input for both arrays

2. Get Input for Arr1 and Arr2

This initializes count to 0

3. Initialize count to 0

This iterates through Arr2

4. For i in Arr2

This iterates through Arr1 (An inner loop)

4.1 For j in Arr1:

This checks if current element is greater than current element in Arr1

4.1.1 If i > j Then

If yes, count is incremented by 1

4.1.1.1 count = count + 1

This ends the inner loop

4.2 End j loop

Print count and set count to 0

<em>4.3 Print count</em>

<em>4.4 count = 0</em>

End the outer loop

4.5 End i loop

End the algorithm

5. End

6 0
2 years ago
A train travels 650 meters in 25 seconds. What is the train's velocity?
frosja888 [35]

The train is traveling 26 meters A second .

3 0
2 years ago
Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph
kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.  

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0
3 years ago
The chart shows the bids provided by four engineers to test a prototype.
klasskru [66]

Answer:

D

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete.  You have to multiply them to get the total cost of the project.

\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]

As you can see, Greg is the least cost-effective because he charges the most for the project.

8 0
3 years ago
Can I get an answer to this question please
crimeas [40]

Answer:

  (i) 12 V in series with 18 Ω.

  (ii) 0.4 A; 1.92 W

  (iii) 1,152 J

  (iv) 18Ω — maximum power transfer theorem

Explanation:

<h3>(i)</h3>

As seen by the load, the equivalent source impedance is ...

  10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

  (36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

<h3>(ii)</h3>

The load current is ...

  (12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

  P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

<h3>(iii)</h3>

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

  (600 s)(1.92 J/s) = 1,152 J

__

<h3>(iv)</h3>

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

  ((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0
2 years ago
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