Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x. find the x component of the velocity of the center of mass (vcm)x at that moment. keep in mind that, in general: vx=dx/dt. express your answer in terms of m1, m2, v1x, and v2x.

Asheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200oc and is permitted to achieve a steady-state diffusion condition. the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10- 11 m2 /s, and the diffusion flux is found to be 1.2 x 10-7 kg/m2 -s. also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3 . how far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3 ? assume a linear concentration profile.

In order to do work, the force vector must be question 1 options: in a different direction than the acceleration vector. in a different direction than the displacement vector. in the same direction as the displacement vector and the motion. in the same direction as the acceleration vector.

Amass m = 74 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius r = 18.4 m and finally a flat straight section at the same height as the center of the loop (18.4 m off the ground). since the mass would not make it around the loop if released from the height of the top of the loop (do you know why? ) it must be released above the top of the loop-the-loop height. (assume the mass never leaves the smooth track at any point on its path.) 1. what is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? 2. what height above the ground must the mass begin to make it around the loop-the-loop? 3. if the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? 4. if the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.4 m off the ground)? 5. now a spring with spring constant k = 15600 n/m is used on the final flat surface to stop the mass. how far does the spring compress?