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3 years ago

Show from the first principles that, for a perfect gas with constant specific heat capacities

Engineering

1 answer:

iVinArrow [24]3 years ago

5 0

The given question is incomplete as the equation given is not making sense.

From internet source i have taken the complete question and its been attached below.

The solution to the question is also attached in next image with complete detail.

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The SDS for any chemical used at a job site must be available

NISA [10]

Answer:

<em>Hello There The Correct Answer is </em>A. At The Job Site.

Explanation:

Because the SDS must not used any kind of chemical Most chemicals used in the workplace have some hazard potential, and thus will be covered by the rule. ... Chemical manufacturers, importers, and distributors of hazardous chemicals are all required to provide the appropriate labels and material safety data sheets to the employers to whom they ship the chemicals. So That why is Opinion A.

3 0

3 years ago

Read 2 more answers

A source which can be represented by a 12 V rms voltage source in series with a resistance of 1.5 k is connected to a 75- load r

Margarita [4]

Answer:

Number of turns ratio is 0.22 and power is 98mw.

Explanation:

Please look at the attached solutions.

4 0

4 years ago

A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

4 years ago

The design product will be a description of the most efficient thermodynamic cycle required to supply 750 MW of electric power.

vlabodo [156]

Answer:

I'm 50s I be sliding on them crabs and we be posting 5 inta a Snapple

3 0

2 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago