Recent content by JDHalfrack

Oh yeah... Duh. It's like I can think of so many things, and something as obvious as this would escape my mind.Thanks for your feedback. This helps my planning so much!

OKay, so this may be a weird place to post this question, but I value the opinion of the members here whether they are current or former students or current or former teachers. I'm setting up an astronomy class for the high school I work at, but unfortunately the way the schedule works is that...

Excellent. That's what I got too. And, think about it. A baseball traveling 45 m/s comes to rest in only 11 cm! It takes a tremendous amount of force to stop it!

Clearly, the question is how can you find acceleration? Hint: if the ball is initially traveling at 45 m/s, and the ball comes to rest after 11 cm, what equation can you use to find acceleration?

That's the same answer I got too. Good job! Glad to help!

Yeah, I'm an idiot. Thank you. The idea was right, I just confused myself. Whoops! Thanks for the correction. I fixed it.

Since you are given the distance over which the force was applied, you should probably use an equation that has distance. The way I read this question, you are given the following: \Deltax = .497m vi = 0 (<-- not given, but I'm assuming the ball wasn't moving until he hit it) vf = 53.6...

Okay, all you found was how far it went in the FIRST FIVE seconds. The question (at least the way your worded it) wants to know how far it goes DURING THE FIFTH second. Do you see the difference? Refer to my previous post if you are confused. If you are still confused after reading that, let...

You're welcome. I'm glad I helped!

Use your kinematics. You know the following: vi = 33.5 km/hr a = -0.29 m/s2 "During the fifth second" means you start at 4 seconds and end an instant before the 5th second. So, you need to find the \Deltax at t = 4 and t = 5. Now, what you do with that knowledge is up to you. I...

Sorry for the confusion. To keep the elevator from accelerating upward or downward (basically to keep it at a constant velocity), the tension force MUST be equal to the force of gravity of the elevator. In essence, the cable is providing an upward acceleration of 9.8 m/s2 to balance out the...

Correct. In fact, whenever the elevator is not accelerating (whether it's stationary or not), the forces are equal because anet = 0 (HINT for #3). Correct again! Now, the key is how much MORE tension? Again, remember Fnet = manet. Since you're now getting the elevator to accelrate upwards...

Sorry for the double post, but here is the question from one of my old textbooks: A rocket is launched at an angle of 53o above the horizontal with an initial speed of 75 m/s, as shown in Figure 3-31 (not included). It moves for 25 s along its initial line of motion with an acceleration of...

Here are a couple hints (I hope they help): - If the elevator was in free fall (accelerating downwards only due to gravity), the cable would not be exerting any force on the elevator. - Also, if the elevator was just hanging there (not going up or down), the cable would have to exert how...

Just a point of clarification, are you sure "after a certain point, begins to fly at a constant speed of 25 m/s for 25 seconds" is correct? I think I have seen this question before (in a book I used to use), and I believe the rocket accelerates at 25 m/s2 for 25 s. But, I may be wrong. It...