Recent content by cugirl

Homework Statement An insulated cup contains 1kg of water initially at 20 C. 0.50 kg of ice, initially at 0 C is added to the cup of water. The water and ice are allowed to come to thermal equilibrium. The specific heat of ice is 2000 J/kg oC, the specific heat of water 4186 J/kg oC...

Good lord, I think I got it. v=6.47

Yes, I did. I thought that was the theorem. My variable, v, is not in that equation you posted. Based on that equation, what would I be solving for? I went back and used LowlyPion's equation.

Here's what I did but I am still not getting the answer: mgh=0.5mv^2 (KE) + 0.5Iw^2 34.5 = .5Iomega^2 mgh*cos30 = 59.7 PLugging in: 59.7 = 1.1v^2 + 34.5 22.9 = v^2 4.8 = v Not a possible answer on our sheet...

Thank you for the person (Lopina) who drew the FBD. Mine did not look like that, and I was able to solve it!

sorry -- that's what I meant. All the greek letters -- uggh.

Homework Statement A block (starting from rest) with mass m = 4.5 kg slides down a frictionless incline a distance of d = 3m and then slides across a horizontal floor with friction (the coefficient of friction is Fk = 0.52). The speed of the block at the bottom of the incline is v1 = 7m/s...

Fk = friction I have Fk - mg = ma Fk - mg = m*v^2 / R

So, Fμ = 10 cos 30 ?

I did. I have Fn perpendicular to the person and pointing toward the center, down is mg, and up is Fk.

Homework Statement A force of 10N is applied downward at an angle of 30 deg with respect to the horizontal on a block. There is friction between the block and the floor and the block remains stationary. M=2kg (a) What is the magnitude of the normal force acting on the block? (b) The...

Homework Statement At an amusement park, some physics students are on a ride where they stand against the wall in a cylindrical apparatus that begins to rotate around its center. Once they are moving fast enough the floor drops and they “stick” to the wall due to friction. The radius of the...

Whoops, I did have a math error. I got omega = 39.6. But, again, using ω = v/r, my v = 39.6*.2, or 7.9, which is not one of the multiple choice answers on the problem set.

If 1/2*I*ω2 = mgh, I solve for the KE at about 69 and omega at 56. But, using ω = v/r, my v = 56*.2, or 11.2, which is not one of the multiple choice answers on the problem set. I am also still uncertain about solving for the # revolutions the disk has turned through when it reaches the bottom.

Using I * a, I get [.5*4*.15^2]*2.1 = .095 for torque. For work, if I plug that in -- .095*25, I don't get one of the multiple choice answers. [This problem set has multiple choice answers.]