Recent content by Draggu

The Attempt at a Solution Hello, I'm having trouble with energy in rotational motion. The question asks: Consider a uniform rod of mass 12 kg and length 1.0 m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins...

Hi, thanks for replying. I re-did the question and figured out what went wrong. I'm extremely silly... in the net force for the sphere I wrote "2a/5" instead of "12a/5" heh. That was enough to mess everything up! Confirmed, got the same answers using both methods. Thanks guys.

How would I go about doing that? Assuming the velocities are the same, I get 3.01m/s.. which is different than the value from forces. Hmm, so you say to relate them. Which means either vb = vp = vs, or vb = vp + vs?

When I try with the energy conservation method, I get a different number. Perhaps you could offer some insight into this? mgh = ΔKE block + ΔKE pulley + ΔKE sphere mgh = (0.5)(4)(v22-v12) + (0.5)(Ipulley)(ω22-ω12) + (0.5)(Isphere)(ω22-ω12) =(0.5)(4)(v22) + (0.5)(0.5)(1kg)(Rp2)(ω22) +...

Homework Statement 15. A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to block of mass 4.0 kg, as shown. What is the speed of the...

Ah! That's what you meant! I think I understand it now! Thanks for all your help (and you're welcome to those who google this question in the future)

Oh ok, so essentially the calculation (as originally) would be: (2kg)(2.8425m/s^2) = F - (0.58)(2KG)(9.8) F=17.005N I wasn't really sure what you meant by "You should check your work to see if the calculated force accelerates the 6 kg block and toboggan system together at the calculated...

Oh, so because we're using static force, then the block is essentially "stuck" to the toboggan, right? In this case, force would be calculated using both the masses, whereas before we only calculated the masses individually because they weren't really moving as a combined system. In this case...

Hi Jay, thanks for replying. I realized this a bit later after I made the post. Here is my new attempt: mBa = F - μs*mg mTa = μs*mg Get an acceleration of 2.8425m/s^2 Which leads to a force of 17.055 N Reasoning: For the block to stay still on the toboggan, I am just looking for the maximum...

Homework Statement A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on top of the toboggan. The coefficient of static friction μs between the block and the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is 0.48. The block is pulled by a...

Alright, so would this be fair to say? Reaction 2 represents the heat evolved as the sodium ion(s) in sodium hydroxide(s) is displaced with the hydrogen ion(aq), producing water and a sodium ion(aq). It also represents the heat evolved from the dissolved NaOH and the neutralization of water...

That's part of reaction 1, and it's already given to us. "Heat evolved as NaOH is dissolves"

I googled it and even searched through threads on this forum, but I still can't get a solid answer. What I know is H1 + H3 = H2 Why? I don't know. I don't understand why they add up. I also don't understand what I am omitting in the 3rd net ionic equation. In the lab it says to...

Hi, I'm still unsure what I am omitting. Part I: I assume reaction 3 is the energy/heat gained as the H and OH molecules react because they are the only two in the net ionic equation. Part II: Perhaps I'm not understanding the purpose of the net ionic equations other than that if there...

I thought about it and I'm still hazy by what you mean. "both cases heat evolved is just that of the neutralization reaction."