The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.
Answer:
Given that
D= 4 mm
K = 160 W/m-K
h=h = 220 W/m²-K
ηf = 0.65
We know that
![m=\sqrt{\dfrac{hP}{KA}}](https://tex.z-dn.net/?f=m%3D%5Csqrt%7B%5Cdfrac%7BhP%7D%7BKA%7D%7D)
For circular fin
![m=\sqrt{\dfrac{4h}{KD}}](https://tex.z-dn.net/?f=m%3D%5Csqrt%7B%5Cdfrac%7B4h%7D%7BKD%7D%7D)
![m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}](https://tex.z-dn.net/?f=m%3D%5Csqrt%7B%5Cdfrac%7B4%5Ctimes%20220%7D%7B160%5Ctimes%200.004%7D%7D)
m = 37.08
![\eta_f=\dfrac{tanhmL}{mL}](https://tex.z-dn.net/?f=%5Ceta_f%3D%5Cdfrac%7BtanhmL%7D%7BmL%7D)
![0.65=\dfrac{tanh37.08L}{37.08L}](https://tex.z-dn.net/?f=0.65%3D%5Cdfrac%7Btanh37.08L%7D%7B37.08L%7D)
By solving above equation we get
L= 36.18 mm
The effectiveness for circular fin given as
![\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D%5Cdfrac%7B2%5C%20tanhmL%7D%7B%5Csqrt%7B%5Cdfrac%7BhD%7D%7BK%7D%7D%7D)
![\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D%5Cdfrac%7B2%5C%20tanh%2837.08%5Ctimes%200.03618%29%7D%7B%5Csqrt%7B%5Cdfrac%7B220%5Ctimes%200.004%7D%7B160%7D%7D%7D)
ε = 23.52
I think it is liters, cubic meters, or milliliters.<span />
The acceleration of gravity on Earth is 9.8 m/s² downward.
This means that gravity adds 9.8 m/s downward to the speed
of a freely falling object every second.
So after 25 sec, it's falling (25 x 9.8m/s) = 245 m/s faster than
it was falling at the beginning of the 25 seconds.
If it dropped from rest (no speed), then its velocity
after 25 seconds is 245 m/s downward.