In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35degree angle. compressing the spring 18 cm causes the ball to hit the floor 1.5m below the point at which it leaves the spring after traveling 6.0m horizontally.

In a wind tunnel the speed changes as the cross sectional area of the tunnel changes. if the speed in a 6' x 6' square test section is 100 mph, what was the speed upstream of the test section where the tunnel measured 20' x 20'? use conservation of mass and assume incompressible flow. conservation of mass requires that as the flow moves through a path or a duct the product of the density, velocity and cross sectional area must remain constant; i.e., that ova-constant. a model is being tested in a wind tunnel at a speed of 100 mph if the flow in the test section is at sea level standard conditions, what is the pressure at the model's stagnation point? (a) the tunnel speed is being measured by a pitot-static tube connected to a u- tube manometer. what is the reading on that manometer in inches of water? (b) at one point on the model a pressure of 2058 psf is measured. what is the local airspeed at that point?

Consider two metallic rods mounted on insulated supports. one is neutral, the other positively charged. you bring the two rods close to each, but without contact, and briefly ground the the neutral rod by touching it with your hand. show answer correct answer what would be resulting charge (if any) on the initially neutral rod

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).