bsbtstrrbt

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 Recognitions: Homework Help What have you done so far? If you don't show us any work you've been able to do, we can't tell you what you might be misunderstanding.
 erererger

 Quote by SusanC1105 Well I know the answers to just about all of them, because we have discussed them in class, but I'm just unclear about some of the detailed physics explanations behind why the answer are correct. Like #1 - I know that the speed on the way up is the same as the way down, I just don't know how to explain why...

That is incorrect.

The velocity of the rock on the way up at height $$x$$ is the same as the velocity of the rock on the way down at height $$x$$
You can prove this using kinematics ($$v_f^2=v_0^2+2ad$$), or using preservation of energy. But that is irrelevant for this problem, as it deals with just the upwards motion.

The first segment you're asked about is from 0 meters above the ground, to 5 meters above the ground. What is the rock's initial velocity (You can calculate this using the two approaches I cited above) and what is its acceleration? You can easily construct an expression describing the time it takes for it to traverse the 5 meters.

Now consider the second segment, which is from 15 meters above the ground, to 20 meters above the ground. What is the rock's velocity at the start of this motion and what is its acceleration? As with the previous segment, you can construct an expression for how long it takes to traverse the 5 meters, or any distance, really.

I suggest you solve this parametrically, as any exam questions are very likely to have follow-up questions which you will find your parametric solution very useful for solving.

EDIT: The answer in my spoiler is irrelevant. I misread the question. Your reasoning is still incorrect, though.

Spoiler
The velocity of the stone is not the same the way up as it is the way down. The acceleration is constant, that much is true, but the last 5 seconds of its flight are the 5 seconds before it hits the floor (20 meters below the throwing point) and first 5 seconds of its flight are when its first cast upwards. They are VERY different.
The formula that has all the answers in this case is $$v_f^2=v_0^2+2ad$$
Since we know the velocity is symmetrical with respect to the maximum height (The velocity of the stone 5 seconds before reaching the maximum height, is the same as its velocity 5 seconds after reaching the maximum height. The same can be said about its height), we'll make the argument as though our stone started from the maximum height, some $$x$$ meters above the throwing point, which is 20 meters above the ground.

Going by the above formula, when is the stone's velocity greater, in the last 5 seconds of its flight (When it clears the ADDITIONAL 20 meters of the height of the throwing point), or in the first 5 seconds of its flight?

 yjttytytjty

 Quote by SusanC1105 Well I know the answers to just about all of them, because we have discussed them in class, but I'm just unclear about some of the detailed physics explanations behind why the answer are correct. Like #1 - I know that the speed on the way up is the same as the way down, I just don't know how to explain why...
I think its better to explain it in terms of "the direction of acceleration"

 Quote by SusanC1105 Well see, this is why I'm here then! Here are my answers: 2. The apparent weight is lighter than the actual weight because of newtons second law. The true weight and the normal force are acting on the person. So the apparent weight= mg +ma Your final answer is correct, but your reasoning is murky. Make a free body diagram and apply Newton's second law of motion. Your apparent weight is defined by the normal force you exert on the scale (And the scale exerts on you). $$mg-N=ma$$ where $$a$$ is the acceleration of the elevator. From here you can isolate N and compare it with its value when the elevator isn't accelerating. 3.Pulling your arm back will reduce the force because you extend the impulse over a greater period of time, which reduces the impact of the force. This is correct. On a test, I would phrase my answer like this, though: $$J\equiv \frac{dP}{dt} \Delta t$$ In order to catch the ball, we must reduce its momentum to 0. Pulling our arm back would mean we increase the time it takes for the ball to undergo the change in momentum. In doing so, we are reducing the average force it exerts on us. Mathematically speaking, $$J$$ is a known quantity, so if we increase $$\Delta t$$, then the average force must decrease accordingly. 4. I'm not too sure on this one, but I think the ball will come out at a 90 degree angle because it is still influenced by the rotational momentum, but doesn't have the string tension there to keep it on course anymore, so it goes off in a new direction? This is incorrect. Remember Newton's first law of motion, and remember that a change in velocity requires that a force acts on the mass in question. And since velocity is a vector quantity (Well, momentum to be more precise), a change in its direction is as much a change as is a change to its magnitude. Try and relate what the tension does in this motion. What it changes. 5. A. potential energy is added to the pendulum prior to release, the pendulum wants to get back to equilibrium and it will take energy to do so. B. kinetic energy is the greatest just as it swings through the equilibrium point on its way down, because it is assisted by gravity? C. The potential energy is highest when the pendulum is at the top of each swing, I don't know how to explain why though. This is correct, but again your reasoning is murky at best. There are only conservative forces acting on the system. Therefore, $$E_k+U_g=constant$$ At what point is the kinetic energy at its maximum (Hint: The potential energy would have to be at its minimum), and at what point is the kinetic energy at its minimum? (Remember that it depends on the velocity of the mass) 6. The frictional force is necessary to negotiate a curve, and gravitational forces are not acting in the right direction to be able to help the car through the curve. This is incorrect. In order to negotiate a curve, there has to be such a force that the car doesn't slip out of the curve, or into it, right? You can take a curve to the limit where it's a circular arc. This is a very big hint. Look at the forces acting on it, and demand that it be in equilibrium so that it doesn't slide up or down the incline. That means that it stays in motion throughout the curve. 7. The sled can not climb a hill higher than its starting point because it only started with a certain amount of potential energy, and since some is lost to frictional forces, there will not be enough to climb a taller hill. 8. There is an advantage with longer contact time, this increased the impulse and puts a greater force on the bat. The bat speed is what creates the force, which is the other part of the impulse equation. J = FdeltaT. Increasing the bat speed will also increase the impulse, making the ball go further. 9.The airbag does the same thing as moving your hand back when catching the baseball, extends the impulse over a greater period of time, reducing the impact you feel. 10. I had to guess at this one, do you swing the bag of oranges back and forth, constantly changing your center of gravity, causing you to move in the direction the oranges are going? 11. this is a transverse wave because the force you are exerting to create these waves is perpendicular to the direction of the waves. 12. It is possible for 2 waves traveling in the same direction to create a smaller wave, as long as the 2 waves are out of phase with eachother, creating interference and thus a smaller wave is produced. 13. Increased tension on a guitar string will increase the frequency and decrease the wavelength of the standing waves. I'm sure there is an equation that proves this, but I'm not sure what it is. 14. I'm pretty clueless on this one. 15. Object A will be easier to set into rotational motion, but I don't understand the physics behind this. 16. Gravity is acting on the person but what is the other force? I don't think its friction because the person is at rest. Could it be the floor acting on the person or a normal force or something? 17. The velocity the ball is rolling at is not an important factor when calculating the time it takes to hit the floor, because this is only dependent on gravity, and the height that the ball falls from. 18. You will see the lighting before you hear the thunder because light waves travel faster than sound waves. I don't really know what physics explanation he is looking for other than that? 19. Sound waves cannot travel through a vacuum because the waves need air particles to move in, and without them, there would be no sound waves. 20. The bird increases the potential energy of the clam when it takes it from the ground, when the bird drops the clam, gravity is acting to bring the clam back to earth. Then there is a collision between the clam and the rock where they both exert a force on eachother, but energy is always conserved. I feel like I have an idea what the answers are, I just can't recall the equations or laws that explain them.

A bit busy now, will reply to the rest later.
Replies in bold.

 dtyjtjdtj
 Did you ever fugure out number 4,7,10, 12, 15 or 16, i got the rest if you need any?
 thrhrhrth
 Is there any that you need
 srthsrthh
 for 5 i used the PE=mgh and explained that height is the only thing changing. and KE= .5mv^2, the other ones im kinda iffy about. 4,10,15 are the worst.