Recent content by mlostrac

Can I just cancel the V's out and solve like this: R2 + 2000 = 2*(2000*R2)/(2000 + R2) Which looks like it would require the quadratic eqn

Homework Statement Two resistors when connected in series to a 110 V line use one-fourth the power that is used when they are connected in parallel. If one resistor is 2.0 kΩ, what is the resistance of the other? Homework Equations V=IR The Attempt at a Solution v=ir...

Hello, I'm just doing a lab on the discharging of a capacitor. The data collected shows that at first the current is large because the voltage is large, so charge is lost quickly and the voltage decreases rapidly. As charge is lost the voltage is reduced making the current smaller so the...

I know, it doesn't make sense! Lol. Well I originally calculated the heat lost by the water using (Q = mc/\T) Q = 150g x 1.00 cal/g-C x (39.6 - 7.1) = 4875 cals Q = 20 kJ So, 20kJ = mc/\T + mL ?? And everything right of the equal sign are the values for ice (ie: mass= 68 g, c = 0.5...

Q = mc/\T = 0.068kg x 0.5 x (0-7.1) = -0.24 kcal x 4.186 kJ/kcal = -1.01kJ Does that say that the ice lost 1.01 kJ? Should I subtract that value from the 294 value?

Do I need that? And what calculation do I need for that? Is it Q = mc(delta)T?

Oook, I think I need the mass of the ice to determine this don't I? Since the final volume was 218 mL, I subtracted that from the initial volume of 150mL to give me 68 mL. Then since water's density is 1g/mL, I used .068kg for the above equation and came out with a value of 294 kJ/kg for L. Is...

I'm not sure? I put that based on the idea that I started with 150mL of water and since water's density is 1g/mL, 150 grams or .150 kg Is that not right?

Hello again, I did a lab where ice was added to 150 mL of warm water and the temperature was observed and recorded. (The mass of the ice was not important; but after it all melted the total volume of the water in the cup was 218 mL) Here's the question I have, "Assuming that the paper cup...

http://img835.imageshack.us/img835/976/exp2pic06.jpg That's the set up. The water is originally at room temp, and then ice and salt are added to the bowl surrounding the glass bottle

Oops, sorry.No, I mean the initial height of the water before it is cooled (when it's at room temperature)

Homework Statement Hello, I'm doing a lab on the Thermal Expansion of Water. I went through the whole lab and got good data based on the the proposed theory however I didn't record the initial height of the water in the pipette. The set up looked like this...

Oook, my bad. I calculated my value for g to be 9.04 which is a little off, but is definitely closer than my first attempt. Thanks so much for all your help!

I'm not even sure if I calculated that right. I just used the equation T=(2π/√g)L1/2, where √9.8 = 3.1 I guess I did that wrong? What does it mean when they want to know the value of g based off the curved fit equation?

I don't think the linear fit graph really matters that much since my values match the curved fit! Thanks for everything. I really appreciate it! One last thing, when we're looking at the curved fit equation, is the value for gravitational acceleration 3.1? What does this say compared to the...