Recent content by BooGTS

Wow, I was way over complicating that, thanks so much RTW69! Then I took the specific volume X the mass of the water and got the volume (just in case someone looks at this in the future).

T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} Found that, but don't really know what the gammas represent?

Homework Statement 1 kg of H20 goes through a reversible adiabatic proces from 220kPa, 200C to 50C in the final state. What is the pressure final and volume final? Assume S1=s2 Homework Equations Conservation of Energy The Attempt at a Solution Found S to be 7.5080, was...

anyone? *bump*

Thanks nrqed, Thats basically what the hint said, but it helped. Here is the finished problem: ((340/340-Vs)*((340+Vs/340)) = 1.164 (115600+340Vs)/(115600-340Vs) = 1.164 115600+340Vs = 1.164 (115600-340Vs) 115600+735.76Vs = 134558.4 735.76Vs = 18958.4 =25.77 m/s I have...

Two pipes, equal in length, are each open at one end. Each has a fundamental frequency of 473 Hz at 297 K. In one pipe the air temperature is increased to 302 K. If the two pipes are sounded together, what beat frequency results? I think I missed this day entirely due to another project. I...

Problem goes as follows: A person standing close to a railroad crossing hears the whistle of an approaching train. He notes that the pitch of the whistle drops as the train passes by and moves away from the crossing. The frequency of the distant approaching whistle is 546 Hz; it drops to 469...

^ yeah, that was it. I followed a similar example to a "T" too...

so, which of the cm are you converting? the area

390 is thermal conductivity (J/s*m*deg C) redoing with "UNITS" would be 390J/s*m*deg C*(2cm^2/100cm/m)*97degK

...im getting it wrong. Calculate the heat flow through a 1.000m long copper bar with a cross sectional area of 2.00 cm2. One end at 2.0 degrees Celsius and the other end at 99.0 degrees Celsius. Rate of heat flow is Q/t. In this problem the coefficient of copper is 390. P=(ka...

I worked on this problem forever before posting and figured it out in this little window on this site of all places. I'm sure it won't be my last tonight though... thanks for the response. 1. 1.85 x10^5 N/m^2 = (x/.0237m^2) 2. 1.85 x10^5 N x .0237 = 4384.5 N = x 3. 4384.5 N x 4 = 17538N =x

Tire inflation to determine weight - solved The problem goes as follows: he four tires of an automobile are inflated to a gauge pressure of 1.85E+5 Pa. Each tire has an area of 0.0237 m2 in contact with the ground. Determine the weight of the automobile. But I've applied the Pfluid = F/A...