While skiing , Sam flies down hill and hits a jump. He has a mass of 75 kg. And he leaves the jump at 18m/s what is his momentum as he leaves the jump

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).

Aplane flies a distance d from west to east at a constant speed vp , with respect to the air in a time twe . on the return trip, the plane flies a distance d from east to west at a constant speed vp , with respect to the air in a time tew . on both trips the wind blows from west to east at a constant speed va , with respect to the ground 1) what is tew in terms of vp, va, and d, as needed? tew=dvp tew=dvp+va tew=dva tew=dvpâ’va tew=12dvp tew=12dvp+va tew=12dvpâ’va your submissions: b submitted: saturday, january 26 at 6: 49 pm feedback: feedback will be given once all questions have been attempted and the grade cluster button has been pressed. 2) what is twe in terms of vp, va, and d, as needed? twe=dvp twe=dvp+va twe=dva twe=dvpâ’va twe=12dvp twe=12dvp+va twe=12dvpâ’va your submissions: d submitted: saturday, january 26 at 6: 49 pm feedback: feedback will be given once all questions have been attempted and the grade cluster button has been pressed. 3) assuming d = 2300 miles, vp = 400 miles/hr, and twe = 4 hours, what is va, the speed of the wind with respect to the ground? va = 100 miles/hr va = 575 miles/hr va = 200 miles/hr va = 175 miles/hr va = 975 miles/hr 4) once again, assuming d = 2300 miles, vp = 400 miles/hr, and twe = 4 hours, what is tew, the time it takes the plane to fly a distance d miles from east to west? tew = 5.75 hours tew = 11.5 hours tew = 7.67 hours tew = 8 hours tew = 10.2 hours

Air initially at 0.75 bar, 1000 k, and occupying a volume of 0.12 m^3 undergoes two processes. process 1-2: the air is compressed isothermally until the volume is halved. process 2-3: the air undergoes a constant pressure process until the volume is halved again. assume ideal gas behavior. a) determine the mass of the air, in kg. b) the work and the heat transfer for each of the two processes, in kj. (100 kj = 1 bar . m^3)