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spin [16.1K]
3 years ago
14

Which is not an affect of alcohol

Engineering
1 answer:
Gre4nikov [31]3 years ago
6 0
Pooping problems is not an affect
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Emergency plans are being formulated so that rapid action can be taken in the event of an equipment failure. It is predicted tha
chubhunter [2.5K]

Answer:

Ammonia gas a hazardous gas to our health, when we are exposed to it for a long time. The gas is lighter than air, that means it's high concentration may not be noticed at the point of leakage, because it flows with the wind direction. Ammonia gas detector are used to determine the concentration of the gas at a particular place. We can use the dispersion modelling software program to know the exact position, where we can place the gas detector, which would be where evacuation is needed.

During evacuation, when the concentration of the gas has increased, a self-contained breathing apparatus should be used for breathing, and an encapsulated suit should be worn to prevent ammonia from reacting with our sweat or any other chemical burn. A mechanic ventilation will also be needed in the place of evacuation, so that the ammonia concentration in that area can be dispersed.

7 0
3 years ago
Read 2 more answers
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

4 0
3 years ago
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

                   =28.12 \ g/mol

3 0
3 years ago
X cotx expansion using maclaurins theorem.
Lemur [1.5K]

It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.

<h3>What is Maclaurin Expansion?</h3>

The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

<h3>What is the explanation for the above?</h3>

as indicated above, the Maclaurin infinite series expansion is given as:

F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

If F(0) = Cot 0

F(0) = ∝ = 1/0

This is not definitive,

Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.

Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
#SPJ1

4 0
2 years ago
A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is
Talja [164]

Answer:

the torque capacity is  30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × \pi × coefficient of friction × maximum pressure ( R³ - r³ )    .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and  maximum pressure in equation 1 and we will get here torque

torque = 2/3 × \pi × 0.2 × 1500 ( 4.5³ - 3.5³ )

so the torque =  30316.369 lb-in

3 0
3 years ago
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