Recent content by KD

Oh, good catch, thanks!

Homework Statement A 7.45g bullet has vi = 353 m/s. It strikes the 0.725 block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 12.1 cm, what was the velocity of the bullet as it emerged from the block? Homework Equations...

Okay, I get this problem now. I was making it way more complicated. The spike can be ignored until the final mass and velocity combined witht eh hammer. Thanks for your help, what you said made exact sense. Thanks!

Would the initial potential energy mgh, have a mass of the spike? Since no distance is mentioned, I'm assuming it is neglible. And then the KEi would be 1/2mv2 with the mass of hammer. + Ui. And that equals mgh with mass of spike? If so, there is no potential energy. Then + .5mv2 with a new...

Ui is 1/3 KEi, right? And KEi is 1/2mv2. But that mass would only be for the hammer. And would the PEf be zero?

Yeah, I got 3.07, along the same answer as you, so yes, it did reach 3.04m. Glad I could help.

A worker drives .5 kg spike into a rail tie with 2.50 kg hammer. Hits spike with 65 m/s. 1/3 kinetic energy converted to internal energy. Find increase in total internal energy. The problem I have having with this problem is that I don't know how to deal with the masses. I know PEi + KEi +...

For this problem, a diagram would be best. Point a would be ball at 2.15m (at rest). Make this your reference leve. And then there is point b, some point higher than point a, assuming its the apex. KEa + PEa = KEb + PEb. Now, there is no potential energy at the reference level so there is only...

We never really went over hydrates, but since some vocab words were thrown at me, the decomp would just be an anhydrous solid and water.

What is the general formula for the decomposition of a hydrate?

I'm fairly sure that if the experiment went according to theory, the graph would be a straight line. I'm saying this because we did a lab like this, and all of our graphs were supposed to be straight lines to show a direct relationship.

To solve this problem, it is best to set up a chart and draw a picture. We already know the time is 1.30s and the accleration is 9.81 m/s2 - that's always a given. To get the vertical height, you need a vertical initial velocity. You need to break the velocity they give you into components...

OK, I feel like the answer is right there, but I'm still not getting it. Why is the centripetal force only applied when it is stretched? Just because it is restraining the object?

First of all, thanks for your responses. And yes, I did mean SPRING, I accidentally typed string the rest of the times. Okay, let me see if I at least kind of got this. Basically the object moves in circular motion because the spring is the force on the object towards the center of the circular...

A heavy object is attached to a spring and then it is twirled in a horizontal ciricle. Why does/does not the string stretch and discuss in terms of force that maintains circular motion. My answer is that the string will stretch because in the horizontal circle motion, the acceleration will...