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den301095 [7]
3 years ago
9

A hydraulic power unit is sometimes referred to as a ?

Engineering
1 answer:
Assoli18 [71]3 years ago
5 0

Answer:

power washer

Explanation:

a hydraullic power unit is sometimes refferd to as a power washer.

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A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicat
Brums [2.3K]

Answer:

The pressure that the water exerts on the bottom of the tank is 59.2 kPa

Explanation:

Given;

height of tank, h = 5m

height of water in the tank, h_w = 4m

pressure at the top of the tank, P_{top} = 20 kPa

The pressure exerted by water at the bottom of the tank is the sum of pressure on water surface and pressure due to water column.

P_{bottom} = \gamma h + P_{top}\\\\P_{bottom} = (9.8*10^3*4 \ \ + \ 20*10^3)Pa\\\\P_{bottom} = 59200 \ Pa\\\\P_{bottom} = 59.2 \ kPa

Therefore, the pressure that the water exerts on the bottom of the tank is 59.2 kPa

6 0
3 years ago
Blank is common during exercise.
dsp73

Answer:

Sweat

Explanation:

As you exercise you respire and warm up due to energy. In turn, two things happen, blood vessels vasodilate (irrelevant to you) and sweat glands sweat more. this sweat then evaporates and cools down the body.

7 0
2 years ago
Which two statements about professional technical jobs in the energy industry are correct?
Tanya [424]
The answer is both B and D
4 0
3 years ago
Read 2 more answers
As a general rule of thumb, in-line engines are easier to work on than the other cylinder arrangements.
siniylev [52]

Answer:

The general rule of thumb is that the SMALLER a substance's atoms and the STRONGER the bonds, the harder the substance. Two of the strongest forms of chemical bonds are the ionic and covalent bonds.

Explanation:

5 0
3 years ago
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0
3 years ago
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