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bonufazy [111]
4 years ago
7

Mining is an example of this type of business

Engineering
1 answer:
luda_lava [24]4 years ago
5 0

Answer:

Mining would go under Industry organization.

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What is 12 cm theodolite​
Klio2033 [76]

Answer:

Answer below :)

Explanation:

<u>SIZE OF THEODOLITE:</u> A theodolite is designated by diameter of the graduated circle on the lower plate. The common sizes are 8 cm to 12 cm while<em> 14 cm</em> to <em>25 cm</em> instrument are used for triangulation work.

4 0
3 years ago
Read 2 more answers
Linus is using a calculator to multiply 5,426 and 30. He enters 5,426 x 300 by mistake. What can Linus do to correct his mistake
Alexxx [7]

Answer:

<em>Linus needs to take one of the zero out and it should be 30 instead 300.</em>

Explanation:

It is because Linus put three zero instead two zero.

4 0
3 years ago
Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.
levacccp [35]

Answer:

\dot W_{out} = 3863.98\,kW

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

h_{in} = 3353.1\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mixture)

h_{out} = 890.1\,\frac{kJ}{kg}

The power produced by the turbine is:

\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})

\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )

\dot W_{out} = 3863.98\,kW

8 0
3 years ago
an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the sur
mel-nik [20]

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.

5 0
2 years ago
The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The abso
ser-zykov [4K]

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

8 0
4 years ago
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