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alina1380 [7]
2 years ago
6

Water enters a vertical jet with low velocity and a pressure of 350 kPa. What is the maximum height that the water can rise abov

e the jet
Engineering
1 answer:
Lapatulllka [165]2 years ago
4 0

The maximum height that the water can rise above the jet is: 35 meters.

<h3>Maximum height that the water can rise above the jet</h3>

Given:

P = 350 kPa = 350000 Pa

We would use pressure(p)  formula to determine the maximum height that the water can rises above the jet by solving for h (height).

Using this formula

Pressure(P) = P₀ + ρgh

Where;

P₀ represent Pressure at the fluid's surface

ρ represent Density of the fluid = 1000 kg/m³

g represent acceleration due to gravity = 10 m/s².

h represent height

Solving for h (height)

350000 = 0 + (1000 × 10 ×h)

350000 = 10000h

Divide both side by  10000h

h = 350000/10000

h = 35 meters

Therefore the maximum height that the water can rise above the jet is: 35 meters.

Learn more about maximum height here:brainly.com/question/28088812

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Answer: Steel is an alloy of iron with typically a few percent of carbon to improve its strength and fracture resistance compared to iron. Many other additional elements may be present or added. Stainless steels that are corrosion and oxidation resistant need typically an additional 11% chromium.

Explanation:

3 0
3 years ago
A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar
jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading  114.06 m^3/d/m

Explanation:

calculation for single clarifier

sewag\  flow Q = \frac{12900}{2} = 6450 m^2/d

surface\  area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2

surface area = 314.16 m^2

volume of tankV  = A\times side\ water\ depth

                             =314.16\times 2 = 628.32m^3

Length\ of\  weir = \pi \times diameter of weir

                       = \pi \times 18 = 56.549 m

overflow rate =v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2

Detention timet_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr

weir loading= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m

6 0
3 years ago
Single point cutting tool removes material from a rotating work piece to generate a cylinder is called • Facing Tuming • Both 1
madam [21]

Answer:Turning

Explanation: Turning is the process in which the work piece is subjected to machining so that excess part is removed with the help of rotation by turning machine or lathe machine.The cutter tool is used for cutting the excess of the work piece and it  is mostly single-pointed so that give accurate removal of the excess of work piece.At times , according to the requirement multi-pointed tool is also used Therefore, the correct option is turning.

6 0
3 years ago
How do you identify all sensors, functions, and where we can use them?
Alex17521 [72]

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7 0
3 years ago
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
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