Recent content by dhkdeoen

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    Calculate humidity from dry/Wet bulb temp

    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...
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    How to Identify the Phase of Steam Using a Thermodynamics Table?

    Homework Statement [/B] Determine the phase of the steam (compress liquid, saturated liquid, saturation mixture, saturated vapor, superheated vapor, or supercritical vapor). Use A-4 Table. T=110 Celsius P=55.732 kPa Homework Equations A-4 Table...
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    Finding the Ratio of Dimensions for Equal Hydrostatic Force on a Vertical Plate

    whoa.. it is hard to get back to old thread. Thanks!
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    Finding the Ratio of Dimensions for Equal Hydrostatic Force on a Vertical Plate

    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...
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    Finding the Ratio of Dimensions for Equal Hydrostatic Force on a Vertical Plate

    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...
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    Getting currents in ammeter, with some resistances.

    I don't think I've heard of that. My class just started with Kirchhoff's two rules.
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    Getting currents in ammeter, with some resistances.

    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!
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    Getting currents in ammeter, with some resistances.

    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..
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    Getting currents in ammeter, with some resistances.

    I just realized I've mixed up my assumptions. i_A+i_topleft=0.5i is this right? and, yes i_A is the current through the ammeter.
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    Getting currents in ammeter, with some resistances.

    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...
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    Getting currents in ammeter, with some resistances.

    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...
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