Recent content by format1998

I need to find the magnitude of the Electric Field at point P. The charge on the left (-50 nC) is 20 cm from the origin, point P is another 5 cm to the right of the origin. So, the -50 nC charge is 25 cm from point P.

Homework Statement (Please see attachment) Point P is 5.00 cm from the origin and each charge is 20.0 cm from the origin. Q1= -50.0 nC (20.0 cm from the origin on the -x axis) Q2= +30.0 nC (20.0 cm from the origin on the +x axis) Point P is 5.00 cm from the origin (in the +x axis)...

Thank you both for your help! Much appreciated!

Q(aluminum) = -10098 J Q (water from 18°C to 0°C) = - 150696 J Q (water to ice phase change) = 666000 J Q (ice from 0°C to -15°C) = -6300 J (-10098 J) + (-150696 J) + (666000 J) + (-6300 J) = 35,132.4 J = 35.13 kJ What am I doing wrong? Are the signs wrong?

"For the Al I used temp change = 33 (18 to -15)" Q = mcΔT 33 grams for the mass of the Aluminum tray? QAl= (.34 kg) (900 \frac{J}{kg*C°})(-15°C - 18°C) did I lay that out wrong or did I use the wrong numbers? "for the water 18" Water in liquid form is from 18°C to 0°C "ice 15"...

Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice Homework Statement 200 g Water (l) is contained in an Aluminum ice tray that has a mass of 340 g. Both is at 18°C. How much heat (Q) must be removed to turn the water into ice at -15°C? Aluminum m[FONT="Times...

So, did I do it correctly? It did say that "to a gas" . From my understanding of the problem as worded, I have to account for the amount of heat required to bring it to its boiling point, as well as the amount of heat required to turn it into a gas, right? Or did I do it incorrectly and I'm...

Homework Statement How much heat is required to change 1.75 L of Ethyl Alcohol (C2H6O) at -50.0°C to a gas at its boiling point? Ethyl Alcohol V = 1.75 L Ti = -50°C 46 g/mol density = 0.789 g/cm3 boiling point = 78°C specific heat (c)= 2400 J/kg*C° Heat of Vaporization (Lv) =...

R = 0.0821 \frac{L*atm}{mol*K} is the value that is on the book and other tables I found on the net Using the mass of one molecule of O2 gave me 479 m/s. One digit off but I'll take it or maybe I'm still doing something wrong?? Thank you

Homework Statement 0.280 mol of O2 gas is at a pressure of 3.50 atm and has a volume of 1.93 L. What is the rms speed (vrms) of the gas molecules? O2 gas n = 0.280 mol @ 32 g/mol m = 0.00896 kg P = 3.50 atm V = 1.93 L Homework Equations PV=nRT -> T = PV/nR...

Thank you to both of you :)

this may be a stupid question but i have to ask as my book doesn't specifically say, it'll help me figure out a problem in my homework. So, here it goes... When an ideal gas expands or is compressed, does the number of moles change or does it remain the same? if it remains the same, does that...

Even though some of the gas leaked out into the environment, the final Volume of the gas remains as 0.08 m3? Is this what you mean by knowing the final Volume?

Homework Statement A tank with a fixed volume of 0.0800 m3 is filled with an ideal gas at a pressure of 4.00 atmospheres and a temperature of 48.0° C. Due to a small leak, some of the gas leaks out. Later it is found that the pressure in the tank is 3.20 atmospheres when the temperature is...