Now that you’ve verified that you know how to make a measurements from this video. Let’s analyze both carts. Use the roller and stopwatch to measure the velocity of each car for the first 30 cm of motion immediately after their hands are no longer in contact. 3. Calculate the momentum of Christine and her cart Asher rules immediately after they are no longer in contact. p=mv 4. Calculate the momentum of Deamon and his cart as he rules immediately after they are no longer in contact. 5. Calculate the total war mentum of the system comprising The two cards and the two people. Data given: Christines cart has a mass of 60 kg Daemon cart has a mass of 66.5 kg The carts have a mass of 22 kg each The velocity closest to daemons cart is V= 1.2 m/s Measured with a ruler at 30 cm apart the time is 0.8875 s.

Asmall, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely responds when it is released? it will move toward the large ball because like charges repel. it will move toward the large ball because like charges attract. it will move away from the large ball because like charges repel. it will move away from the large ball because like charges attract.

Suppose water is leaking from a tank through a circular hole of area ah at its bottom. when water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to cah 2gh , where c (0 < c < 1) is an empirical constant. a tank in the form of a right-circular cone standing on end, vertex down, is leaking water through a circular hole in its bottom. (assume the removed apex of the cone is of negligible height and volume.) (a) suppose the tank is 20 feet high and has radius 8 feet and the circular hole has radius 2 inches. the differential equation governing the height h in feet of water leaking from a tank after t seconds is dh dt = − 5 6h3/2 . if the height of the water is initially 8 feet, how long will it take the tank to empty? (round your answer to two decimal places.)

Arocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay = 12.80 m/s32t, where the +y-direction is upward. (a) what is the height of the rocket above the surface of the earth at t = 10.0 s? b) what is the speed of the rocket when it is 325 m above the surface of the earth?