This equation is known as the ideal gas law, and it can be used to predict the behavior of many gases at relatively low pressure. from this equation, you can see that as the temperature of a gas increases, a) the pressure of the gas increases. b) the number of moles of gas will go down. c) either the pressure of the gas, the volume of the gas, or both, will increase. d) either the pressure of the gas, the volume of the gas, or both, will decrease.

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).

When you jump from an elevated position you usually bend your knees upon reaching the ground. by doing this, you make the time of the impact about 10 times as great as for a stiff-legged landing. in this way the average force your body experiences is a) less than 1/10 as great. b) more than 1/10 as great. c) about 1/10 as great.d) about 10 times as great.

Aspecified volume of space contains an electric field for which the magnitude is given by e=e0cos(ωt). suppose that e0 = 20 v/m and ω = 1.0 × 107 s−1. part a what is the maximum displacement current through a 0.80 m2 cross-sectional area of this volume? express your answer with the appropriate units. idisp,max i d i s p , m a x = nothing nothing request answer part b how would this area have to be oriented relative to the electric field to get this maximum displacement current?