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3 years ago

Where does Mr. Teller work? What do they do there?

Engineering

1 answer:

natulia [17]3 years ago

7 0

Most business owners begin his business in order to increase profit and to expand.

<h3>What are Business Practices?</h3>

This refers to the various ways in which a business owner decides to organize his business and the policies which guides it.

With this in mind, we can see that Mr X believes that it is a good business practice to <em>prioritize the work</em> that seems the most difficult and the most likely to kill their projects but this is not a good business practice because it can put the entire business in jeopardy.

Please note that your question is incomplete so I gave you a general overview so that you could get a better understanding of the concept.

Read more about business practises here:
brainly.com/question/1343903

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Water at 20 °C is flowing with velocity of 0.5 m/s between two parallel flat plates placed 1 cm apart. Determine the distances f

Basile [38]

Answer:

The distance from the entrance at which the boundary layers meet is 0.516m

The distance from the entrance at which the thermal boundary layers meet is 1.89m

Explanation:

For explanation, look at the attached file

3 0

3 years ago

Can the United States defeat Iranian forces

Lostsunrise [7]

Answer:

yes for sure

Explanation:

iran oy has 3000 ballistic missiles but america has thousands of nukes. we can easily outnumber their forces as well. this will be a easy victory for us but will result in massive casualties on both sides

5 0

3 years ago

Read 2 more answers

Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by

irakobra [83]

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

5 0

3 years ago

Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.

saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 $\mathbf{m^{3}}$

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C ; $v_1 = vg_1$ = 0.0996 $\mathbf{m^{3}}$ / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 $\mathbf{m^{3}}$ / kg

From temperature T₂ = 200⁰ C

$v_f_2 = 0.0016 \ m^3/kg$

$vg_2 = 0.127 \ m^3/kg$

Since $vf_2 < v_2$ , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality $x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}$

$x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}$

$x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}$

$\mathsf{x_2 =0.78}$

At temperature T₂, the specific internal energy $u_f_2 = 850.6 \ kJ/kg$ , also $ug_2 = 2594.3 \ kJ/kg$

Thus,

$u_2 = uf_2 + x_2 (ug_2 -uf_2)$

$u_2 =850.6 +0.78 (2594.3 -850.6)$

$u_2 =850.6 +1360.086$

$u_2 =2210.686 \ kJ/kg$

<u>At state 3:</u>

Temperature $T_3=T_2 = 200 ^0 C ,$

$V_3 = 2V_1 = 0.06 \ m^3$

Specific volume $v_3 = 0.2 \ m^3/kg$

Thus; $vg_3 =vg_2 = 0.127 \ m^3/kg$ ,

SInce $v_3 > vg_3$ , therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at $v_3 = 0.2 \ m^3/kg$ and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 $\ m^3/kg$

The specific internal energy $u_3$ at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

$u_2-u_1$

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

$u_3-u_2$

= (2622.3 - 2210.686) kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0

3 years ago

For a short time a rocket travels up and to the left at a constant speed of v = 650 m/s along the parabolic path y=600−35x2m, wh

julia-pushkina [17]

Answer:

Detailed working is shown

Explanation:

The attached file shows a detailed step by step calculation..

4 0

3 years ago