What is the area of 5 meter by 16 meter sheet of wood?

A0.17kg ball rolls at 0.75m/s to the right on a frictionless surface and collides with a 0.17kg ball rolling to the left at 0.65m/s. (elastic collision) what is the final velocity of the second ball when the first is 0.5m/s to the left? what is the change in kinetic energy before and after?

During physical science, ms. greene challenged her students to produce an energy transformation. james and jill wrapped a 4-inch nail with a coil of fifty turns of wire. they connected one end of the coil to one terminal of a knife switch. they connected the other terminal of the knife switch to the battery. finally, they connected the end of the coil to the other terminal of the battery. james held the tip of a 1-inch nail near the flat end of the 4-inch nail. jill closed the knife switch quickly and then opened it. the 1-inch nail was pulled toward the 4-inch nail. what is the best explanation of why the nail moved? a) electrical energy was converted to mechanical energy in the 1-inch nail. b) the 4-inch nail became an electromagnet and the magnetic force attracted the 1-inch nail. c) the electric current in the 4-inch nail was converted to mechanical energy in the 1-inch nail. d) the 4-inch nail had a positive charge and the 1-inch nail had a negative charge. opposites attracted.

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).