Answer:
Q = 5 L/s
Explanation:
To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):
![Q=\frac{V}{t}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7BV%7D%7Bt%7D)
V: Volume (volumen) = 200L
t: time (tiempo) = 40 s
you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):
![Q=\frac{200L}{40s}=5\frac{L}{s}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B200L%7D%7B40s%7D%3D5%5Cfrac%7BL%7D%7Bs%7D)
Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)
It's a quantitative observation because it includes numerical data.
Answer:
As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.
Explanation:
Depression in freezing point :
![\Delta T_f=K_f\times m](https://tex.z-dn.net/?f=%5CDelta%20T_f%3DK_f%5Ctimes%20m)
where,
=depression in freezing point =
= freezing point constant
m = molality ( moles per kg of solvent) of the solution
As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:
- If molality of the solution in high the depression in freezing point of the solution will be more.
- If molality of the solution in low the depression in freezing point of teh solution will be lower .
Relative lowering in vapor pressure of the solution is given by :
![\frac{p_o-p_s}{p_o}=\chi_{solute}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_o-p_s%7D%7Bp_o%7D%3D%5Cchi_%7Bsolute%7D)
= Vapor pressure of pure solvent
= Vapor pressure of solution
= Mole fraction of solute
![p_s\propto \frac{1}{\chi_{solute}}](https://tex.z-dn.net/?f=p_s%5Cpropto%20%5Cfrac%7B1%7D%7B%5Cchi_%7Bsolute%7D%7D)
Vapor pressure of the solution is inversely proportional to the mole fraction of solute.
- Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
- lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.
Your answer is the ball it's very simple a ball rolling down hill will be the one that makes the most sense.
Answer:
young's modulus
Explanation:
We have given length of tendon L= 13 cm =0.13 m
Change in length ![\Delta L=3.7mm=3.7\times 10^{-3}m](https://tex.z-dn.net/?f=%5CDelta%20L%3D3.7mm%3D3.7%5Ctimes%2010%5E%7B-3%7Dm)
Force = 12.1 N
Average diameter d =8.8 mm
So ![r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B8.8%7D%7B2%7D%3D4.4mm%3D4.4%5Ctimes%2010%5E%7B-3%7Dm)
Area ![A=\pi\times (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ctimes%20%20%284.4%5Ctimes%2010%5E%7B-3%7D%29%5E2%3D60.79%5Ctimes%2010%5E%7B-6%7Dm%5E2)
Now stress ![=\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bforce%7D%7Barea%7D%3D%5Cfrac%7B12.1%7D%7B60.79%5Ctimes%2010%5E%7B-6%7D%7D%3D0.1990%5Ctimes%2010%5E6N%2Fm%5E2)
Strain ![=\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bchange%5C%20in%5C%20lenght%7D%7Blength%7D%3D%5Cfrac%7B3.7%5Ctimes%2010%5E%7B-3%7D%7D%7B0.13%7D%3D28.46%5Ctimes%2010%5E%7B-3%7D)
Now young's modulus