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.
Answer:
0.0659 A
Explanation:
Given that :
( saturation current )
at 25°c = 300 k ( room temperature )
n = 2 for silicon diode
Determine the saturation current at 100 degrees = 373 k
Diode equation at room temperature = I = Io ![\frac{V}{e^{0.025*n} }](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7Be%5E%7B0.025%2An%7D%20%7D)
next we have to determine the value of V at 373 k
q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1
Given that I is constant
Io =
= 0.0659 A
Answer:0.1898 Pa/m
Explanation:
Given data
Diameter of Pipe![\left ( D\right )=0.15m](https://tex.z-dn.net/?f=%5Cleft%20%28%20D%5Cright%20%29%3D0.15m)
Velocity of water in pipe![\left ( V\right )=15cm/s](https://tex.z-dn.net/?f=%5Cleft%20%28%20V%5Cright%20%29%3D15cm%2Fs)
We know viscosity of water is![\left (\mu\right )=8.90\times10^{-4}pa-s](https://tex.z-dn.net/?f=%5Cleft%20%28%5Cmu%5Cright%20%29%3D8.90%5Ctimes10%5E%7B-4%7Dpa-s)
Pressure drop is given by hagen poiseuille equation
![\Delta P=\frac{128\mu \L Q}{\pi D^4}](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cfrac%7B128%5Cmu%20%5CL%20Q%7D%7B%5Cpi%20D%5E4%7D)
We have asked pressure Drop per unit length i.e.
![\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20P%7D%7BL%7D%20%3D%5Cfrac%7B128%5Cmu%20%5C%20Q%7D%7B%5Cpi%20D%5E4%7D)
Substituting Values
![\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20P%7D%7BL%7D%3D%5Cfrac%7B128%5Ctimes8.90%5Ctimes10%5E%7B-4%7D%5Ctimes%5Cpi%20%5Ctimes%5Cleft%20%28%200.15%5E%7B3%7D%5Cright%20%29%7D%7B%5Cpi%5Ctimes%204%20%5Ctimes%5Cleft%20%28%200.15%5E%7B2%7D%5Cright%20%29%7D)
=0.1898 Pa/m
Answer:
The answer is "
".
Explanation:
Please find the correct question in the attachment file.
using formula:
![\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \ or \ \ \frac{RT_1 -RT_2}{n-1}\\\\](https://tex.z-dn.net/?f=%5Cto%20W%3D-P_1V_1%2BP_2V_2%20%5C%5C%5C%5CWhen%20%5C%5C%5C%5C%5Cto%20W%3D%20%5Cfrac%7BP_1V_1-P_2V_2%7D%7Bn-1%7D%5C%20%5C%20%20%20or%20%5C%20%5C%20%20%5Cfrac%7BRT_1%20-RT_2%7D%7Bn-1%7D%5C%5C%5C%5C)
![W =\frac{R(T_1 -T_2)}{n-1}\\\\](https://tex.z-dn.net/?f=W%20%3D%5Cfrac%7BR%28T_1%20-T_2%29%7D%7Bn-1%7D%5C%5C%5C%5C)
![=\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.287%2825%20-237%29%7D%7B1.5-1%7D%5C%5C%5C%5C%3D%5Cfrac%7B0.287%28-212%29%7D%7B0.5%7D%5C%5C%5C%5C%3D%5Cfrac%7B-60.844%7D%7B0.5%7D%5C%5C%5C%5C%3D-121.688%20%5Cfrac%7BKJ%7D%7BKg%7D%5C%5C%5C%5C%3D-121%20%5Cfrac%7BKJ%7D%7BKg%7D%5C%5C%5C%5C)