Homework Statement
Dry-bulb Temperature 46'C
Wet-bulb Temperature 34'C
P=101.3kPa
with these, I have to 'calculate' Specific Humidity(w), Relative humidity(φ), Specific Enthalpy(h)Homework Equations
Specific Humidity
w=0.622*Pv/Pa
when Pa=P-Pv
Relative Humidity
φ=w*P/[(0.622+w)*Pg]
Specific...
so..
er.. so I divide both side with R^2
(L/R)^2=pi*L/(2R)+2/3
since L/R=x
x^2=pi/2x+2/3
x^2-pi/2x-2/3=0
and this gives me two roots
x1=1/12(3pi-sqrt(96+9pi^2)=-0.3475...
x2=1/12(3pi+sqrt(96+9pi^2)=1.91832..
so I pick the one bigger than 0; which is x2.
so L/R should be...
1. A vertical plate consists of rectangular and semicircular portion and has dimensions as shown. It is submerged in a liquid such that the upper edge coincides with the free surface of the liquid. What is the ratio of L/R such that force on the rectangular portion is the same as that on the...
it is! Maybe I was too hanging out with R1. Now I got the right answer.
Even though I got the right answer, I really should trying to get answer with R1, though. So much appreciated for your devotion!
then it should be
i_A=0.5i-(((R*R1)/(R+R1)+R/2)*i-0.5iR)/(1.29R))
substitute R1=1.29R
i_A=0.5i-(1.06332iR-0.5iR)/(1.29R)
i_A=0.5i-(0.56332/1.29)i
i_A=0.5i-0.43668217i
i_A=0.06331877iI=V/R
i=ε/R
so this should be the answer, but still..
I couldn't managed to pull out the answer, so I did what you said.
R_t=(R*R1)/(R+R1)
R_b=R/2
R_eq=(R*R1)/(R+R1)+R/2
ε=R_eq*i
ε=V_t+V_b=V_t+0.5iR
V_t=R_eq*i-0.5iR
i_topleft=V_t/(1.29R)
so
i_A+i_topleft=i
i_A=i-i_topleft...
Homework Statement
In the figure R1 = 1.29R, the ammeter resistance is zero, and the battery is ideal. What multiple of ε/R gives the current in the ammeter?
Homework Equations
V=IR, 1/R=1/R_1+1/R_2..(when resistances are parallel)
etc
The Attempt at a Solution
since ammeter doesn't...