H2so4 goes to hso4 did it gain or lose a proton? is it a bronsted-lowery acid or base?

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

5submission this assignment is worth 20 points total. you are required to submit the following by next lab: 1. (3 points) determine the equation for the output angular velocity ω2 = θ˙ 2 as a function of θ1, ω1 = θ˙ 1 and α. you must show all your work to receive credit. 2. (2 points) use the result of problem#1 to plot ω2 over 0 ≤ θ1 ≤ 360deg with ω1 = 360deg/sec. do this for α = {10,30}deg. show the results on the same plot and properly label the axes, title, legend. for this you can use matlab or ms excel. 3. (3 points) determine the equation for the output angular acceleration ω˙ 2 and create a plot similar to the one in problem#2. 4. (10 points) submit plots of the results (ω2 and ω˙ 2) obtained from creo/mechanism and compare them to results of problem#2 and problem#3. note that you should do these comparisons for the two cases with α = {10,30}deg. 5. (2 points) provide a brief explanation of the results. did they match? what are the implications as misalignment angle increases?

Abasketball star covers 2.65 m horizontally in a jump to dunk the ball. his motion through space can be modeled precisely as that of a particle at his center of mass. his center of mass is at elevation 1.02 m when he leaves the floor. it reaches a maximum height of 1.90 m above the floor and is at elevation 0.910 m when he touches down again. (a) determine his time of flight (his "hang time"). (b) determine his horizontal velocity at the instant of takeoff. (c) determine his vertical velocity at the instant of takeoff. (d) determine his takeoff angle. (e) for comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.45 m, and yf = 0.750 m.