Recent content by liz_p88

Thank you! I doubt myself so much. I have to work so hard to understand these concepts and formulas

Homework Statement A 2.50 meter long copper rod has a mass of 0.25 kilograms. A magnetic field of 0.15 Teslas is oriented horizontally from south to north. a.) What is the minimum current in the rod that would allow it to "levitate" in this magnetic field? b.) What is the direction of the...

Homework Statement On a day when the air temperature is 19 °C, a 0.15 kg baseball is dropped from the top of a 24 m tower. After the ball hits the ground, bounces a few times, and comes to rest, by how much has the entropy of the universe increased? Homework Equations ΔS = Q/T Also gravity...

That's what I figured. Thank you for double checking for me.

Homework Statement A high jumper of mass 60.0 kg consumes a meal of 3.00 x 10^3 kcal prior to a jump. If 3.3% of the energy from the food could be converted to gravitational potential energy in a single jump, how high could the athlete jump? Homework Equations Ug = mgh gravity =...

Ok I think I have it. Q gained by the ice = Q lost by the water mLf + mc(Tf - 0) = mc(Ti - Tf) m(333.7) + mc(0-0) = (500)(4.186)(25) = 157 g

Thats where I get confused. Okay so because of latent heat, I would have to use the equation...mcΔt + mL + mcΔt = -[mcΔt]...or Q=mcΔt + mL? This is where I get confused. The only thing is, I don't know how to plug in the variables. I know it takes 333.7 kJ to change 1 kg of ice to water at 0 celcius

Homework Statement Ice at 0°C is mixed with 5 x 10^2 mL of H2O at 25°C. How much ice must melt to lower the H2O temp to 0°C? Homework Equations mcΔt = -[mcΔt] The Attempt at a Solution (500ml)(4.186 kJ/kg*K)(-25) = m(2.1 kJ/kg*K)(25) 996 ml or gram equivalents I know this...

Homework Statement A particle with a charge of 3.0 µC is located at rest 5.0 meters from a proton. a) Determine the magnitude of the electric field at the proton's location created by the 3.0 µC charge. b) Determine the initial acceleration of the proton after it is released from rest...

I figured it out. (101 kPA)(.045 m^3)/303k = (2020 kPA)(.004995 m^3)/T2 673k or 400°C

Homework Statement A cylinder in a car engine takes Vi = 4.50 x 10^-2 m^3 of air into the chamber at 30°C and at atmospheric pressure. The piston then compresses the air to one-ninth of the original volume (0.111Vi) and to 20.0 times the original pressure (20.0 Pi). What is the new...

I did it again, this time I did square root of (2)(5.6511 x 10^-21)/ (4.648 x 10^-26) 493 m/s

Homework Statement Find the rms speed in air at 0.0°C and 1.00 atm of (a) the N2 molecules (b) the O2 molecules, and (c) the CO2 molecules. Homework Equations vrms = square root of 3kT/m The Attempt at a Solution (a). 28.0u x (1.66 x 10^-27) = 4.68 x 10^-26 1.5 x (1.38 x 10-23...

Awesome thank you for your input!

That makes sense. So would I only have to account for the temperature increase and disregard the contraction when solving for (a)? And did I calculate it correctly?