If anyone has any complaints/thinks im worng/is confused by my answers, please email me at janovi@wm.edu . I'll respond as quickly as I can.
1.) A dolphin can leap several meters above the ocean's surface. Why doesn't gravity stop the dolphin from leaving the water?
Answer:The dolphin is able to supply a force greater than gravity's and by the 2nd law, leap into the air. Also, you could say that when the dolphin reaches the surface, it has an intertial tendency to jump into the air. However, gravity will pull it down.2.) As you jump across a small stream, does a horizontal force keep you moving forward? If so, what is that force?
Answer: No, there is no horizontal force on your body while you are moving forward. When you first leap, you orient your body so you'll push off the ground at an angle, which means you'll push the ground vertically and horizontally. By Newton's 3rd Law, the ground will push back vertically and horizontally. Once you leave the ground, however, the reaction forces produced by the ground vanish, and the only force acting on your body is gravity.8.) Why is your velocity continuously changing as you ride on a carousel?
Answer: Because your position is constantly changing and velocity is the CHANGE in position. The velocity vector is tangental to your location. Some time later, let's say you moved about a quarter circle. The velocity vector at this time is now pointing in a completely different direction. So essentially, the components change while the speed remains constant.16.) An acorn falls from a branch located 9.8 m above the ground. After 1 second of falling, the acorn's velocity will be 9.8 m/s downward. Why hasn't the acorn hit the ground?
Answer:Even though after 1 s the acorn is moving 9.8 m/s, its average veloctiy over the interval of 1 s is 4.9 m/s so it has only moved half the distance.26.) A car passes by, heading to your left, and you reach out and push it towrad the left with a forceof 50 N. Does this moving car push on you and, if so, with what force?
Answer: Yes, the car will push on you. By Newton's 3rd Law, when you push on the car, the car will exert a reaction force to the right with magnitutde of 50 N. But of course, this force will last only a moment since the car is moving and your hand will have little time to remain keep contact
28.) Comic book superheroes often catch a falling person only a hairsbreadth from the ground. Why would this rescue actually be just as fatal for the victim as hitting the ground itself?
Answer: The person of mass m falls from a large height from rest, so just before they hit the ground, they are moving FAST. The superhero has to apply a force equal to mg to STOP THE ACCELERATION. However, by v2 = v1 + a*t, the person is still moving. So the superhero has to add to mg to make the person accelerate upward enough to keep them from going splat. However, the person is REALLY REALLY close to the ground, so t will be VERY small. And since v2 = 0, the eq becomes -v1 / t = a. Since t will be less than one and v1 is the speed of the person just before the superhero makes the save, a will be HUGE, which means the force the superhero has to pull upward with will be so large it will probably kill the person.
8.) A sprinter can reach a speed of 10 m/s in 1 s. If the sprinter's acceleration is constant during that time, what is the sprinter's acceleration?
Answer: We can find the acceleration easily with a kinematic equation. The sprinter's motion is in one dimension, we don't have to worry about components. We will use
11.) If you jump upward with a speed of 2m/s, how long will it take before you stop rising?
Answer: You give yourself an initial velocity of 2 m/s, or v1= 2 m/s. As you are rising in the air, gravity is pulling you down and eventually you stop (v2 = 0) and start descending. So, we know a, v1, v2, and we'll use the same eq as in the previous problem. Plugging away gives you t = 0.204s
20.) You're using a wedge to split a log. You are hitting the wedge with a large hammer to drive it into the log. It takes a force of 2000 N to push the wedge into the wood. If the wedge moves 0.2 into the log, how much work have you done on the wedge?
Answer: We know the work we do on an object is calculated by multiplying the applied force and the distance it travels in the direction of motion. In this case, all the force is along the direction of motion, so the work is (2000N)*(0.2 m) = 400 J