A0.0434-m3 container is initially evacuated. then, 4.19 g of water is placed in the container, and, after some time, all of the water evaporates. if the temperature of the water vapor is 417 k, what is its pressure? number units in pa

Contrast the force of gravity between these pairs of objects a 1 kg mass and a 2 kg mass that are 1 m apart and two 2 kg masses that are 1 m apart

1point) a skydiver weighing 130 lb (including equipment) falls vertically downward from an altitude of 14,000 ft and opens the parachute after 20 s of free fall. assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude 0.85|v|0.85|v| when the parachute is closed and is of magnitude 17|v|17|v| when the parachute is open, where the velocity vv is measured in ft/s. assume that acceleration due to gravity has magnitude 32 ft/s/s; remember that weight is the product of mass and gravitational acceleration. find the speed of the skydiver when the parachute opens. speed is 150.6121788 ft/s find the skydiver's height when the parachute opens. height is 2338.98591 ft find the limiting velocity vlvl after the parachute opens. (think about whether this should be positive or negative.) vl=vl= 7.647 ft/s

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).