Recent content by David89

I ended up getting an A- in the course and I would say that an emphasis on conceptual understanding is FAR MORE important than continuous drilling. In fact, I would say that the continuous drilling prevented me from getting a better grade because when it came time for the exam and my response...

Yes that is where the anomaly is. The bottle was taken from room temp and placed in a container filled with hot water when the temperature inside the bottle reached 15 degrees celsius. How would I include the temperature of the surrounding area as a factor in my graph? I can see what youre...

Okie doke, So I switched the x and y axis, thanks for that suggestion it looks much better that way. I also performed a curve fit to get a function which reads 0.044x^2 - 0.6174x + 4.9. This makes sense, when the temp is 0 the reading is 5.2, so I'd say the function is reasonably accurate...

Homework Statement I've been doing an experiment where I measure the water volume in a glass bottle as I manipulate the temperature from room temp (24.8 degrees celsius) to 0 degrees celsius, and then back up to room temp. I know that water behaves uniquely from 0 to 4 degrees celsius, it...

It's difficult to say, I've only written one final from Athabasca for physics and this one was it. The questions connect areas from the whole course and introduces new questions never dealt with, so definitely bring your brain to the exam and don't expect to simply memorize the info or...

Sure! Once you get t, you can plug t into your equation for V to solve for your initial vertical velocity. Once you have initial vertical velocity, you can resolve your vector into horizontal and vertical components, and you can set the final velocity for vertical motion to zero (since Vf is...

Thanks to everyone who helped in this thread!

It sounds like you did a lot of work much like myself, I watched videos from Khan academy and put more into this course than I ever have for a course before, best of luck tomorrow

Yea, the course goes through this and discusses the benefits of leaving out the variables until the final step. This is my first ever physics course, and I have not done math in 7 years, so I get too excited and plug in too early cause I want the answer right away. I got to learn to be more...

at the risk of making you nervous, I will say that I felt the exam was much harder than the midterms, assignments, and questions on the lab reports. I had 89% in the course going into the exam, and I'm not confident that I got a 65%. I did all the assigned problems twice, the practice problems...

I solved for Vx, Vy, max height, and final velocity as well. I really needed to solve this question. I think it's all good.

Ok, I think this is good

Hmmmm ok. I will rewrite this on paper and post back soon

OHHH, so if I plug in V for the horizontal equation of motion into the vertical equation for motion, 0 = 5500sin39*t/t*cos39 -4.9t^2. The t's in the first term cancel, and sin39/cos39 is equal to tan39, so our equation becomes 0 = 5500tan39 - 4.9t^2. OR -4454 = -4.9t^2 sqrt(-4454/-4.9) =...

Ohhhh, ok, I think I get it, when the vertical displacement is zero, the rock is on the ground. I was changing the reference frame for some reason, putting Y=0 at 1200m elevation so it looked like the equation in post 6. If I solve for t, then 0 = 1200 + 5500/t*cos39 -4.9t^2, at this point...