Answer:
Explanation:
Hello,
In this case, by using the Bernoulli equation:
Whereas 
accounts for the heat loss, so we can compute the change of pressure by:
![p_1 +\frac{1}{2}\rho v^2_1+\rho gh_1= p_2+\frac{1}{2}\rho v^2_2+\rho gh_2+\rho h_L\\\\p_2-p_1=\frac{1}{2}\rho v^2_1+\rho gh_1-\frac{1}{2}\rho v^2_2-\rho gh_2-\rho h_L\\\\\Delta p=\rho *[\frac{1}{2}(v_1^2-v_2^2)+g(h_1-h_2)-h_L ]](https://tex.z-dn.net/?f=p_1%20%2B%5Cfrac%7B1%7D%7B2%7D%5Crho%20v%5E2_1%2B%5Crho%20gh_1%3D%20p_2%2B%5Cfrac%7B1%7D%7B2%7D%5Crho%20v%5E2_2%2B%5Crho%20gh_2%2B%5Crho%20h_L%5C%5C%5C%5Cp_2-p_1%3D%5Cfrac%7B1%7D%7B2%7D%5Crho%20v%5E2_1%2B%5Crho%20gh_1-%5Cfrac%7B1%7D%7B2%7D%5Crho%20v%5E2_2-%5Crho%20gh_2-%5Crho%20h_L%5C%5C%5C%5C%5CDelta%20p%3D%5Crho%20%2A%5B%5Cfrac%7B1%7D%7B2%7D%28v_1%5E2-v_2%5E2%29%2Bg%28h_1-h_2%29-h_L%20%5D)
Thus, we must first compute the velocity inside the 5-cm section by using the continuity equation:
It means that the change in pressure in Pa turns out (density is 1500 kg/m³ given the specific gravity of the fluid):
![\Delta p=1500\frac{kg}{m^3} *[\frac{1}{2}((1.2\frac{m}{s})^2-(1.92\frac{m}{s})^2)+9.8m/s^2(0m-15.0m)-25\frac{m^2}{s^2} ]\\\\\Delta p=1500\frac{kg}{m^3}*(-1.1232\frac{m^2}{s^2} -147\frac{m^2}{s^2} -25\frac{m^2}{s^2} )\\\\\Delta p=-259684.8Pa](https://tex.z-dn.net/?f=%5CDelta%20p%3D1500%5Cfrac%7Bkg%7D%7Bm%5E3%7D%20%20%2A%5B%5Cfrac%7B1%7D%7B2%7D%28%281.2%5Cfrac%7Bm%7D%7Bs%7D%29%5E2-%281.92%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%29%2B9.8m%2Fs%5E2%280m-15.0m%29-25%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%20%20%5D%5C%5C%5C%5C%5CDelta%20p%3D1500%5Cfrac%7Bkg%7D%7Bm%5E3%7D%2A%28-1.1232%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%20-147%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%20-25%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%20%29%5C%5C%5C%5C%5CDelta%20p%3D-259684.8Pa)
Therefore, the change in pressure in bar and MPa turns out:
Such negative sign means that the pressure at the 5-cm section is lower than the pressure at the 8-cm section.
Best regards.
Carbonate compounds have a negative valency (because they contain oxygen), so NaCO3 has a negative valency (it’s a negatively-charged ion) which makes it basic or alkaline.
Answer:
KE = 1/2*m*v^2
KE = 1/2*150kg*(20 m/s)^2
KE = 75kg * 400m²/s²
KE = 30,000 kg*m²/s²
KE = 30,000 N*m
KE = 30,000 J
Explanation:
Hope this helped.
A brainliest is always appreciated.
The molar concentration of a solution containing 469 grams of C6H12O6 in 82 milliliters of water is 31.7M.
<h3>HOW TO CALCULATE MOLAR CONCENTRATION:</h3>
- The molar concentration of a solution can be calculated by dividing the number of moles by its volume.
- According to this question, there are 469 grams of C6H12O6 in 82 milliliters of water.
The number of moles = 469g ÷ 180.156g/mol = 2.6mol
Volume of water in litres = 82/1000 = 0.082L
Concentration of solution = 2.6mol ÷ 0.082L = 31.71M
Therefore, the molar concentration of a solution containing 469 grams of C6H12O6 in 82 milliliters of water is 31.7M.
Learn more about concentration at: brainly.com/question/202460
