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3 years ago

As an employee, who's is supposed to provide training on the chemicals you are handling or come in contact with at work?

Engineering

2 answers:

masha68 [24]3 years ago

5 0

I think it will be C!

Karo-lina-s [1.5K]3 years ago

4 0

C is the right answer

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Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s and e

ZanzabumX [31]

Answer:

exit temperature 285 K

Explanation:

given data

temperature T1 = 270 K

velocity = 180 m/s

exit velocity = 48.4 m/s

solution

we know here diffuser is insulated so here heat energy is negleted

so we write here energy balance equation that is

0 = m (h1-h2) + m × $(\frac{v1^2}{2}-\frac{v2^2}{2})$ .....................1

so it will be

$h1 + \frac{v1^2}{2} = h2 + \frac{v2^2}{2}$ .....................2

put here value by using ideal gas table

and here for temperature 270K

h1 = 270.11 kJ/kg

$270.11 + \frac{180^2\times \frac{1}{1000}}{2} = h2 + \frac{48.4^2\times \frac{1}{1000}}{2}$

solve it we get

h2 = 285.14 kJ/kg

so by the ideal gas table we get

T2 = 285 K

4 0

3 years ago

Is a compass a analog or a digital sensor?

ollegr [7]

A compass is a analog sensor

5 0

2 years ago

A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

podryga [215]

Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

$T_{H} = 1.967\,^{\circ}F$

b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5\cdot T_{H} - 2398.35 = 479.67$

$T_{H} = 575.604\,R$

$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

$T_{H} = 298.279\,K$

$T_{H} = 25.129\,^{\circ}C$

8 0

2 years ago

If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B

Sergio039 [100]

Answer:

105 km

Explanation:

The motorist was going 30 km/hr, and it took 3 hours 30 minutes. That's 3.5 hours. 3.5×30=105

5 0

3 years ago

What do you think of web 3.0? do you think it will be realized someday in the future?

Elodia [21]

Answer:

is this a question for hoework

7 0

2 years ago