I don't know if you guys can clearly see the image attached. (sorry i don't have my good digitial camera with me here). But the question is:
A 0.5 kg block B slides without friction inside a slot cut in arm OA which rotates in a vertical plane at a constant rate, (dΘ /dt) = 2 rad/s. At the...
I don't know if you guys can clearly see the image attached. (sorry i don't have my good digitial camera with me here). But the question is:
A 0.5 kg block B slides without friction inside a slot cut in arm OA which rotates in a vertical plane at a constant rate, (dΘ /dt) = 2 rad/s. At the...
An object of unknown mass is hung on the end of an unstretched spring and is released from rest. If the object falls 3.42 cm before first coming to rest, find the period of the motion.
Doing a summation of forces, i see that the force constant k times the displacement y is equal to the...
well the summation of forces on the cube shows the tension force is equal to the weight of the cube minus the Buoyant force. The new forces in the beaker system are the Normal force by the balance pan and the weight of the beaker. But isn't the weight of the beaker dependent on the weight of...
ok i see. thank you, Doc Al, for helping me out. I spent hours on that problem and didnt realize that it would be like that. I know the problem is done and all, but how do you know that the buoyancy force is equal to the weight of the mass on the right pan? I mean to say, what principles or...
so are you saying the force mg on the right pan is equal to the buyoyant force on the cube in the beaker on the left pan? If this were true, then the density of the water times the volume of the cube would equal that mass m; which would turn out to be 0.064 kg. Is this correct? Or is there...
A beaker filled with water is balanced on the left pan of a balance. A cube of 4 cm on an edge is attached to a string and lowered into the water so that it is completely submerged. The cube is not touching the bottom of the beaker. A weight of mass m is added to the right pan to restore...
Ok, so my first integral would be \int (-mg) dr with limits of integration being 0 to Re. I add this to the second integral which is \int (-GMem/r^2) dr with limits of integration being from Re to \infty.
So, my total potential energy would be -2mgRe.
So at the Center of the Earth, the...
hey thanks, so the integral would contain [(-GMem)/r]dr with limits of integration being Re and Infinity? I guess i have a hard time seeing why this would give me the escape speed from the center of the earth...Wouldnt taking the integral of the Force of gravity show how much work is being...
How would one go about calculating the escape velocity of an object with mass m from the center of the Earth. I understand that that when launched from the surface of the Earth, mechanical energy is conserved and you end up with v escape =sqrt[2gRe] So what i did was i calculated what i...