What is the placement of a lock out device on an energy isolating device?.

We would like to measure the density (p) of an ideal gas. We know the ideal gas law provides p= , where P represents pressure, R

Nostrana [21]

Answer: =

Explanation:

= P / (R * T) P- Pressure, R=287.058, T- temperature

From the given that

Sample mean(pressure) = 120300 Pa

Standard deviation (pressure) = 6600 Pa

Sample mean(temperature) = 340K

Standard deviation(temperature) = 8K

To calculate the Density;

Maximum pressure = Sample mean(pressure) + standard deviation (pressure) = 120300+6600 = 126900 Pa

Minimum pressure = Sample mean (pressure) - standard deviation (pressure)= 120300-6600 = 113700 Pa

Maximum temperature = Sample mean (temperature) + standard deviation (temperature) = 340+8 = 348K

Minimum temperature = Sample mean (temperature) - standerd deviation (temperature) = 340-8 = 332K

So now to calculate the density:

Maximum Density= Pressure (max)/(R*Temperature (min))= 126900/(287.058*332)= 1.331

Minimum density=Pressure(min)/(R*Temperature (max))= 113700/(287.058*348)= 1.138

Average density= (density (max)+ density (min))/2= (1.331+1.138)/2= 1.2345

cheers i hope this helps

5 0

3 years ago

When designing a car that runs on wind or Air car . can you tell me the details for the following points Compressed Air Engine:

BabaBlast [244]

Answer:

a)

The crack and connecting rod is used in the design of car.This mechanism is known as slider -crank mechanism.

Components:

1.Inlet tube

2. Wheel

3. Exhaust

4. Engine

5.Air tank

6.Pressure gauge

7.Stand

8. Gate valve

b)

The efficiency of air engine is less as compare to efficiency of electric engine and this is not ecofriendly because it produce green house gases.These gases affect the environment.

c)

it can run around 722 km when it is full charge.

5 0

3 years ago

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125

Anna [14]

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

4 0

1 year ago

What is the force in kN of work done is 1.2 ms moves through 120m

Semmy [17]

Answer:

$\frac{1.2}{120}$

0.01

5 0

2 years ago

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

3 years ago