Recent content by huskydc

How can you separate the series and parallel in this combination?? I'm confused... the question asks for equivalent resistance I thought it's somelike like this: 1/R eq = 1/2R(1) + 1/R(1) + 1/R(2) but it's wrong...help?

:frown: :frown: can i even use the equation: delta U = 1.5 n R delta T but that would include finding out n...which i don't have enough info to calculate...any one help?? i'm lost... =(

A cylinder (cross section is 0.2m2) with a free moving piston is filled with gas. The piston is attached to a heavy weight W = 10000N. Outside the cylinder, the air is at 300K and 1atm. Initially the gas is at 300K, then it is heated to 400K. The heat capacity of the gas under the constant...

ok...that works...i was thinking since it's heat gained by ice cubes...i'll use the specific heat for ice... then may i ask under what kind of situation do i use specific heat for ice?? conditions where..ice...is still in temp... below zero??

ok...so... Q water = 200 (4.186)(9) = 7534.8 Q ice = m (334 + (2.093 x 16)) = 367.488 m Q water = Q ice 7534.8 / 367.488 = m of ice = 20.5 kg... since each cube weighs .03 kg... 20.5/.03 = 684 cubes? i tried that didn't work...unless the units I'm...

i'm sorry folks, I'm really lost with this one? can someone help me out?

but the problem is I'm not sure how to set it up... i try something like this, and i know it's wrong... Q water =7534800 J each ice cube is .3 kg, Q ice = m c delta T: .3(2090)(16) = 10032 J melting? Q = 200 x (33.5 e 4) = 6.7 e 7 i'm just confused...

During one hot summer, a physics grad student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, he filled it with 200 liters of water at 25 C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each...

ok, after much try, i got total height difference to be 2.3 m. now it's asking: How much did the height of the car drop when the person got in the car? which i think it's asking how much of that height difference is attributable to either side. Because the areas of the sides are...

I've been told to do this : deltaP = rho*g*h solve for h (P=F/A, so solve for F/A on each side. The difference in these two is deltaP, then divide by rho*g) i tried that didnt work, or is it right but I am just doing it wrong? :cry:

Here's how i got 63.5 kg in the very first part: we are in an enclosed cylinder...(well hydraulic lift), if i push down on piston 1 with F(1), it increases the pressure in that cylinder by : change in pressure = F(1)/ A(1) Eqa. 1 by pascal's principle, the pressure in...

well, i guess my problem is first of all what it is really asking... is it asking for the change in height of the piston (fluid) when the person get into the car? b/c that doesn't make sense since that's the next part of the problem...

oops, I'm sorry, i couldn't find my original post, i thought it got deleted somehow...sorry about that...

A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 630 cm2. It is filled with oil of density 690 kg/m3. I've found that approx. 63.5 kg of mass must be placed on the small piston to support a car of mass 1600 kg at equal fluid levels. but then it asks...

A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 630 cm2. It is filled with oil of density 690 kg/m3. I found that about 63.5 kg of mass must be placed on the small piston to support a car of mass 1600 kg at equal fluid levels. But then it asks... With...