The diagram shows a transistor used in a circuit. A signal with a small change in voltage is input into a circuit that includes a transistor connected to 4 resistors and 2 capacitors. The signal output from the transistor has a much larger change in voltage than the input.
What does the diagram show?

an NPN transistor used as an amplifier
an NPN transistor used as a switch
a PNP transistor used as an amplifier
a PNP transistor used as a switch

The slotted link is pinned at o, and as a result of rotation it drives the peg p along the horizontal guide. compute the magnitude of the velocity and acceleration of p along the horizontal guide. compute the magnitudes of the velocity and acceleration of p as a function of θ if θ=(3t) rad, where t is measured in seconds.

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?

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).