Answer:
Is there supposed to be an attachment?
Explanation:
Answer:
1.25 cm/day
Explanation:
An air thickness , (l) = 0.15 cm
Air Temperature =
![(T_a)=20^0C = (20+273)K\\(T_a)=293K](https://tex.z-dn.net/?f=%28T_a%29%3D20%5E0C%20%3D%20%2820%2B273%29K%5C%5C%28T_a%29%3D293K)
Mass Diffusion coefficient (D) = ![0.25cm^2/sec](https://tex.z-dn.net/?f=0.25cm%5E2%2Fsec)
If the air pressure ![(P_a) = 0.5 P_{sat}](https://tex.z-dn.net/?f=%28P_a%29%20%3D%200.5%20P_%7Bsat%7D)
We are to determine how fast will the water
level drop in a day.
From the property of air at T = 20° C
from saturated water properties.
The mass flow of
can be calculated as:
![H_2O = \frac{D}{\phi} \delta C](https://tex.z-dn.net/?f=H_2O%20%3D%20%5Cfrac%7BD%7D%7B%5Cphi%7D%20%5Cdelta%20C)
where:
![\delta C = \frac{P_{sat}*P_a}{RT }](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%20%5Cfrac%7BP_%7Bsat%7D%2AP_a%7D%7BRT%20%7D)
R(constant) = 8.314 kJ/mol.K
![\delta C = \frac{2.34*0.5}{8.314*293 }](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%20%5Cfrac%7B2.34%2A0.5%7D%7B8.314%2A293%20%7D)
![\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%204.803%2A10%5E%7B-4%7D%5C%5C%20%5Cdelta%20C%20%3D0.48%2A10%5E%7B-3%7D%20mol%2Fm%5E3%5C%5C%20%5Cdelta%20C%20%3D%200.48%2A10%5E%7B-6%7D%20mol%2Fcm%5E3)
Since 1 mole = 18 cm ³ of water
will be: ![(0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3](https://tex.z-dn.net/?f=%280.48%2A10%5E%7B-6%7Dmol%2Fcm%5E3%20%2A18%29cm%5E3%2Fcm%5E3)
![\delta C = 8.64 * 10^{-6}](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%208.64%20%2A%2010%5E%7B-6%7D)
Again:
![H_2O = \frac{D}{\phi} \delta C](https://tex.z-dn.net/?f=H_2O%20%3D%20%5Cfrac%7BD%7D%7B%5Cphi%7D%20%5Cdelta%20C)
![= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.25%7D%7B0.15%7D%2A8.69%2A10%5E%7B-6%7D%20%5Cfrac%7Bcm%5E2%2Fsec%7D%7Bcm%7D)
![=1.4481*10^{-5} \frac{cm^2/sec}{cm}](https://tex.z-dn.net/?f=%3D1.4481%2A10%5E%7B-5%7D%20%5Cfrac%7Bcm%5E2%2Fsec%7D%7Bcm%7D)
Converting the above value to cm/day: we have:
![1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}](https://tex.z-dn.net/?f=1.448%2A10%5E%7B-5%7D%2A3600%2A24%5Cfrac%7Bcm%7D%7Bs%7D%2A%5Cfrac%7Bs%7D%7Byr%7D%2A%5Cfrac%7Byr%7D%7Bday%7D)
= 1.25 cm/day
∴ the rate at which the water level drop in a day = 1.25 cm/day
Answer:
If it is rechargable, there should be instructions on it, but I believe you can recycle it. For single use batteries, they need to be put into a safe location as they can actually explode sometimes, and acid leaks can also occur.
Normally, you can use put something like electrical tape or a nonconductive material on top of both of the battery nodes and you can just toss it into a trash can.
Explanation:
Answer:
speed(0)
square_length = int(input("What is the length of the square?"))
def draw_square():
pendown()
for i in range(4):
forward(square_length)
left(90)
penup()
penup()
setposition(-200, -200)
draw_square()
penup()
setposition(-200, 200 - square_length)
draw_square()
penup()
setposition(200 - square_length, 200 - square_length)
draw_square()
penup()
setposition(200 - square_length, -200)
draw_square()Explanation: