Look at picture 1 and 2. Answer my question correctly.

Askydiver weighing 140 lb (including equipment) falls vertically downward from an altitude of 19,000 ft and opens the parachute after 18 s of free fall. assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude 0.55|v| when the parachute is closed and is of magnitude 14|v| when the parachute is open, where the velocity v 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. (a) find the speed of the skydiver when the parachute opens. (b) find the distance fallen before the parachute opens. (c) what is the limiting velocity vl after the parachute opens? (d) determine how long the sky diver is in the air after the parachute opens. (e) plot the graph of velocity versus time from the beginning of the fall until the skydiver reaches the ground.

Which of the following is not a means to accelerating? question 4 options: a)increase speed b)remain still c)decrease speed d)change direction

Amass m = 74 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius r = 18.4 m and finally a flat straight section at the same height as the center of the loop (18.4 m off the ground). since the mass would not make it around the loop if released from the height of the top of the loop (do you know why? ) it must be released above the top of the loop-the-loop height. (assume the mass never leaves the smooth track at any point on its path.) 1. what is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? 2. what height above the ground must the mass begin to make it around the loop-the-loop? 3. if the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? 4. if the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.4 m off the ground)? 5. now a spring with spring constant k = 15600 n/m is used on the final flat surface to stop the mass. how far does the spring compress?