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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...

well, the book has pretty much the same example, it says to find the final speed of the clay and platform system, find kinetic energy of the system after collision then set this KE to .5kA^2, tried that example myself, it didnt work out ...

ok, the figured out the k = 98.1 N/m, next the problem asks: With what amplitude does the platform oscillate immediately after the clay hits the platform? i first used the momentum conservation, and found the velocity of the clay+platform system is equal to .36 since total E =...

A platform of mass 0.8 kg is supported on four springs (only two springs are shown in the picture, but there are four altogether). A chunk of modeling clay of mass 0.6 kg is held above the table and dropped so that it hits the table with a speed of v = 0.9 m/s. The clay sticks to the table...

so... I alpha = Torque (1/12) (10.6)(6.8^2) + (5.29)(3.4^2) alpha = 3.4 * (5.29 * 9.81) sin37 solve for alpha? (well i already did, b/c I'm getting a little desperate..and its wrong) :frown:

ok, here's my take on total of moment of inertia: 1/12) (5.29)(6.8^2) + (10.6)(3.4^2) ? and torque of mass: 3.4 * (5.29 * 9.81) sin37 ?? i'm sorry, I'm a microbiology major, I'm really really bad at physics... i really thought i had the right answer, i mean I've figured out...

i'm getting myself more confused than ever, the net torque is just of the torque exerted by mass itself. the moment of inertia for rod is (1/12) M L^2. the rod's concentrated mass is still at the axis. so: total moment of inertia: (1/12) (5.29)(6.8^2) -- mass do i add (1/12)...

so would the net torque be just of the torque exerted by mass?

A uniform rod is pivoted at its center and a small weight of mass M = 5.29 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. Suppose that the rod has length L = 6.8 m and mass mrod = 10.6 kg. Suppose also that...

hold on...it says it's rotating about its symmetry axis, doesn't that make I = 3/2 m r^2? anyways...if we assume I to be .5 m r^2, i calculate the KE like this: KE initial = .5 I(1) omega(initial)^2 + .5 I(2) omega(initial)^2 = .5 [.5 m(1) r(1)^2] omega(initial)^2...

but it says that friction brings the two disks to a common rotational speed final. shouldn't i be using the same though?

A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at finitial = 152 rpm. A second disk of mass M2 = 247 g and radius R2 = 5 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed...

the answer turns out to be .0857m, I guess we'er all too caught up in the big picture and totally forgot about the omega being 2pi, thanks

=( , i tried them all, none of them works...

0.09 is also incorrect...

A 100 g ball is attached to a rubber tube and is spun around in a circle at a rate of one revolution every second. A force of 0.5 N is required to stretch the tube 1.0 cm. If the original length is L = 1.0 m, what will be the change in length of the rubber tube when the ball is revolving...

yea, i initially thought it's asking for F(A), and i tried solving for it, didnt work out, unless i did something wrong in my original calculations...

A man of mass mm = 95 kg decides to paint his house. To do this, he builds a platform using a uniform beam with a mass of mb = 100 kg and a length of L = 7 meters. The beam is supported by two sawhorses, as shown in the diagram above. If the man stands over the support at point B, calculate...

i got it, thanks!

A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.77. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on...

ok, i got it, thanks!

now we have initial PE = final KE, it goes... .5kx^2 = .5m(1)v(1)^2 + .5m(2)v(2)^2 does it make sense? but the problem now is i have two variables to solve...one equation...

also, wouldn't PE final be zero too? after the carts are released, the spring would be relaxed again, thus no compression, thus zero.. that'll make it: initial PE = final KE but i don't know where to go from here

sorry doc, i really have no clue on this...i know i need to find the speed. since here...the ball rolls w/o slipping. thus the velocity is equal to its tangential velocity.

a tire, rolls up on a ramp, with a speed of 2.8 m/s, a tire mass of 10 kg and tire radius of 30 cm the tire cruises up an embankment at 30° for 1 meter What is the velocity of the tire at the topof the embankment in m/s? I tried the following: energy conservation: and end up...

may i ask what explosion? or collision? the only thing happening is that the cart is dropping off coal..

momentum is conserved, i kept getting 20 m/s does the time, 4 sec matter in answering this question?

A small ball of radius r = 2.4 cm rolls without slipping down into a loop-the-loop of radius R = 2.5 m. The ball has mass M = 352 gm. How high above the top of the loop must it be released in order that the ball just makes it around the loop?

A massless spring of spring constant 20 N/m is placed between two carts. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 3.5 kg. The carts are pushed toward one another until the spring is compressed a distance 1.8 m. The carts are then released and the spring pushes them apart. After the...

ok, I've set up the equations as follow: .5kx^2 = KE (in this case, also KE final) thus, .5kx^2 = .5m(4.963^2) and solve for x... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?

A railroad hopper car has mass 50000 kg when empty and contains 50000 kg of coal. As it coasts along the track at 10 m/s the hopper opens and steadily releases all the coal onto a platform below the rails over a period of 4 s. How fast does the car travel after all the coal is dumped? i...

A projectile of mass 20 g has an initial horizontal velocity of 100 m/s when it hits and stops in a wood block of mass 0.403 kg. The block is sitting on a horizontal frictionless surface and is attached to a massless spring, initially relaxed, with spring constant 148 N/m. What is the maximum...

I have the same problem as Aki, what I did I set initial Energy = final Energy, since energy is conserved, so i end up having : gh = 0.5V^2 , just like what ehild had above. But how I can I get V?

ok, i got it. i found F, and i realized that i have to take the angle at which the spring was stretched into account. Thanks for the help!

ok, i got that part down, its distance max is 4.64 m, but here's another part of the problem: what is the magnitude of the acceleration of the block at this point? (when spring is stretched farthest?) what I did was I tried to identify the force exerted by the spring, find the x comp...

I'm half way through this problem, but I'm stuck at the end. A block of mass m = 4.5 kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = 5 m. The spring has spring constant k = 11 N/m and is relaxed when hanging in the vertical position. The block is...