Recent content by catzmeow

Haha I tried this and the result in the academic world wasn't very pretty...

The problem says that initial velocity is 8 m/s, but I understand what you're saying.

Hi PF, I'm reviewing my notes from class, starting from the very beginning. I'm working on some problems using kinematic equations. Here's one example: a ball falls from 30m (using down as the positive direction) at a velocity of 8 m/s. how long does it take the ball to hit the ground? Why do I...

I want to say I did do that, I can't exactly remember now. I wasn't sure if I included to many changes or not...thanks for your input!

Hmmm I only wish I could go back and change it! Thank you, chestermiller, I'll have to remember this for the next test...

Haha yes very true. For some reason I feel like I shouldn't have included the specific heat of water or of ice... I'm not sure. Something felt off about it but I can't put my finger on it...

Hi PF! I just took a test for my class and there is one problem I can't get my mind off of. I'm not necessarily looking for the correct solution here, just if my thinking was correct. It went more or less like this: there is 40g of ice at 0 degrees C in 200g of water in an aluminum calorimeter...

Oh ok- that makes since. Thanks Barryj!

Oops I meant all that I have equal to mu_k

Thanks barryj! I did that and now I have MA= (Mgsin(theta))/(input force-mgcos(theta)... Is this close(er)?

I'm utterly lost on this and would appreciate any help! I'm given mechanical advantage= 1/(sin(theta)+mu_k*cos(theta)) And efficiency = 1/(1+mu_k*cot(theta)) These are for an inclined plane and theta is not specified, I'm just supposed to show how those equations were derived. Thanks!

yes, but i don't see where the 0.2 (or m) comes in your calculations :confused: You're right- I didn't actually use these to calculate the mu_k, they were part of the experiment to calculate the ideal mechanical advantage and mechanical advantage. Sorry for the confusion :(

My apologies- I mistyped. I meant that we used grams but I had already converted them to Newtons using the gravity constant. So my calculations are done with Newtons. The .2 was the weight in kilograms of the force.

Sure! Sorry :( We measured the force needed to move the a wooden block (2.67 N) up the incline (also wood) for 20, 30 and 40 degrees. Acceleration is constant. For an incline of 20, we used 1.96 N/200 g to move the block. For 30, we used 2.21 N/225 g to move the block For 40, we used 2.45...