C, due to stuff and whatnot
Answer:
1,050 Joules
Explanation:
<u>Step 1:</u> work done in moving the box 30 meters
work done = force X distance
= 25N X 30 = 750 Joules
<u>Step 2: </u>calculate total internal energy
Total internal energy = work done + kinetic energy
= 750 Joules + 300 Joules
= 1,050 Joules = 1.05 KJ
The compound is (Sulphuric Acid) H2SO4. On reacting with (Sodium Hydroxide) NaOH, it gives (2 Water Molecules/Colored) 2H2O and (1 Sodium Sulfate Molecule/Salt) Na2SO4
H2SO4 + NaOH —> 2H2O (aq.) + Na2SO4 (salt)
The resulted salt/compound (Na2SO4) when reacting with Methyl Orange (MO) is called ”Removal of methyl orange dye and Na2SO4 salt from synthetic wastewater using reverse osmosis (RO)”
The efficiency of reverse osmosis (RO) membranes used for treatment of colored water effluents can be affected by the presence of both salt and dyes.
Concentration polarization of each of the dye and the salt and the possibility of a dynamic membrane formed by the concentrated dye can affect the performance of the RO membrane.
The objective of the current work was to study the effect of varying the Na2SO4 salt and methyl orange (MO) dye concentrations on the performance of a spiral wound polyamide membrane.
The work also involved the development of a theoretical model based on the solution diffusion (SD) mass transport theory that takes into consideration a pressure dependent dynamic membrane resistance as well as both salt and dye concentration polarizations.
Control tests were performed using distilled water, dye/water and salt/water feeds to determine the parameters for the model.
The experimental results showed that increasing the dye concentration from 500 to 1000 ppm resulted in a decrease in the salt rejection at all of the operating pressures and for both feed salt concentrations of 5000 and 10,000 ppm.
Increasing the salt concentration from 5000 to 10,000 ppm resulted in a slight decrease in the percent dye removal. The model’s results agreed well with these general trends.
Answer:
P=9.58 W
Explanation:
According to Newton's second law, and assuming friction force as zero:
![F_m=m.a\\F_m=0.875kg*a](https://tex.z-dn.net/?f=F_m%3Dm.a%5C%5CF_m%3D0.875kg%2Aa)
The acceleration is given by:
![a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D%5C%5Ca%3D%5Cfrac%7B0.685m%2Fs%7D%7B21.5%2A10%5E%7B-3%7Ds%7D%5C%5C%5C%5Ca%3D31.9m%2Fs%5E2)
So the force exerted by the motor is:
![F_m=0.875kg*31.9m/s^2\\F_m=27.9N](https://tex.z-dn.net/?f=F_m%3D0.875kg%2A31.9m%2Fs%5E2%5C%5CF_m%3D27.9N)
The work done by the motor is given by:
![W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m](https://tex.z-dn.net/?f=W_m%3DF_m%2Ad%5C%5C%5C%5Cd%3D%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E2%5C%5Cd%3D%5Cfrac%7B1%7D%7B2%7D%2A31.9m%2Fs%5E2%2A%2821.5%2A10%5E%7B-3%7Ds%29%5E2%5C%5C%5C%5Cd%3D7.37%2A10%5E%7B-3%7Dm)
![W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J](https://tex.z-dn.net/?f=W_m%3D27.9N%2A7.37%2A10%5E%7B-3%7Dm%5C%5CW_m%3D0.206J)
And finally, the power is given by:
![P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BW_m%7D%7Bt%7D%5C%5CP%3D%5Cfrac%7B0.206J%7D%7B21.5%2A10%5E%7B-3%7Ds%7D%5C%5C%5C%5CP%3D9.58W)
Answer:
induced current
Explanation:
intentionally manipulated.