Recent content by coldsteel

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    Height of oil of above water U cylinders

    so 2cm is the correct answer?
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    Height of oil of above water U cylinders

    An open U shaped tube is initially partially filled with water (ρ=1000 kg/m^3). A 10 cm high column of oil (ρ=800 kg/m^3) is poured into the right side of the tube. What is the height of the oil above the water? I set the water and oil equal to each other. (800)(10)(9.8)=(1000)(H)(9.8)...
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    Velocity of Water in 5 cm Tube: Solve Equation

    1.A tube of radius 5 cm is connected to tube of radius 1 cm. Water is forced through the tube at a rate of 10 liters/min. The pressure in the 5 cm tube is 1×10^5 Pa. The density of water is 1000 kg/m3. Assume that the water is nonviscous and uncompressible. What is the velocity of the water in...
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    2D momentum piece breaks problem

    1.A piece of ice falls on a frozen lake and breaks up into three pieces which go off in the directions shown in the diagram below. The mass of the bigger piece is 4 kg and it moves with a speed of 4 m/s straight down, If the two smaller pieces are 2 kg each and break off at 45 degrees each. What...
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    Plank and two supports using torque

    Ahhhh...after struggling for 2 hours and wearing down my eraser...I think I may have solved this. take a look at what I did and let me know if its finally right. these are the equations I used. I had 2 unknown so I set the pivot point on the left support to =0. 1.)...
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    Plank and two supports using torque

    the 3m is half the length of the board. so should I move my pivot point to the far left support? so the center mass of the plank would be 3m? This is the equation I get: 50(9.8)(3m)-Force left(0m)+100(9.8)(4m)=0 there's nothing to solve with that though
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    Plank and two supports using torque

    A man with a mass of 100kg is standing on a uniform plank of mass 50kg. The man is standing directly above the right support and the left support is located at the left end of the plank. The right support is 2m from the right end. the plank is 60m long. What is the magnitude of force the left...
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    Failed first test on forces&kinematicsneed some help with corrections

    ok so the other force acting on it would be friction. I used Fnet=ma which m(a) is 0. so it would be T-Friction=0. friction is mu x normal. so i could solve for mu which is the coefficient of friction would is equal to 0.36. I am now stuck on the bucket swug in a vertical cirlce. I used the...
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    Failed first test on forces&kinematicsneed some help with corrections

    OK I'm stuck on 3. So to find the net force on the block A I used Fnet=ma. Since it states its moving at constant velocity, the acceleration =0. So the net force would be 0? Is this correct? What about finding the coefficient of friction if acceleration is 0? I know the force of friction is...
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    Failed first test on forces&kinematicsneed some help with corrections

    unfortunatly I failed my first physics exam. I have to go back through and correct the ones I missed. I need some help on the ones I am stuck on. any help and explanations is greatly appreciated. 1.) three blocks are being accelerated upward by a force applied to the bottom block. the mass...
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