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ra1l [238]
2 years ago
7

The fumes from are extremely hazardous Cement Does cement have fumes

Engineering
1 answer:
IRINA_888 [86]2 years ago
8 0

Answer:

Cement is largely made up of calcium oxide. Cement can harden in the gastrointestinal tract and cause obstruction. Inhaling cement dust can cause coughing, wheezing, and difficulty breathing. The best prevention is to avoid or limit contact with cement.

Explanation:

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An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava
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Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.

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A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass
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2 years ago
Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and
ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, Q=16 m^{3}/min

Diameter d= 8cm= 0.08 m

Assumptions

  • The flow is jet flow hence momentum-flux correction factor is unity
  • Gravitational force is not considered
  • The flow is steady, frictionless and incompressible
  • Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

V=\frac {Q}{A} where V is the exit velocity and A is area

Area, A=\frac {\pi d^{2}{4} where d is the diameter

By substitution

V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min

To convert v to m/s from m/s, we simply divide it by 60 hence

V=\frac {3183.098862  m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s

3 0
2 years ago
Transcript
posledela

Answer:

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Explanation:

8 0
3 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
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