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Lostsunrise [7]
3 years ago
6

Which of the following hazards is shared by surface mining and sub-surface mining?

Engineering
2 answers:
lubasha [3.4K]3 years ago
4 0

B I guess. since they both have potential to collapse

Alex777 [14]3 years ago
4 0

Answer: Hazardous chemicals

Explanation:

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9
Alex17521 [72]

Answer:

do it if it's Sunday why are you like this

6 0
3 years ago
Why do engineers (and others) use the design process?
Naily [24]

Explanation:



Engineering design is an iterative process used to identify problems and develop and improve solutions. The engineering design process can be extremely useful to any individual trying to solve a problem.

4 0
3 years ago
A 20-mm thick draw batch furnace front is subjected to uniform heat flux on the inside surface, while the outside surface is sub
lara [203]

Answer:

hello your question is incomplete attached below is the complete question

<em>answer</em> :

To ( inside temperature ) = 598 K

TL ( outside temperature ) = 594 k

Explanation:

a) Determine the surface temperature To and TL based on the known conditions provided in the drawing

To ( inside temperature ) = 598 K

TL ( outside temperature ) = 594 k

attached below is the detailed solution

5 0
3 years ago
What is the difference between a Datum and a Datum Feature? a) A Datum and Datum Feature are synonymous. b) A Datum is theoretic
aliina [53]

Difference between Datum and Datum feature is<em> 'Datum is theoretical and Datum feature is real'. </em>

Option: (b)

<u>Explanation:</u>

A Datum is a perfect plane, line, point or surface but only occurs theoretically.

However a Datum Feature is fully based on a tangible surface, axis or point on a part where that theoretical datum is located.

The reason behind in this is they are not equal to each other because the 'part surface' is never 100% perfect.

The important functional features of the Datum is controlled during measurements.

7 0
3 years ago
One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p
kupik [55]

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)  

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

3 0
3 years ago
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