Recent content by lmbiango

I want to delete a post or maybe several posts that I have made in the past, or otherwise change my "screen name" because I probably should have picked something a little more incognito. Either way, I can't find a way to do either of these things and I, as a fairly intelligent person, have been...

So, are you saying that the difference between the weights would be zero? I thought that would make sense, but I tried that also (although I thought it seemed too easy) and the homework tool said it wasn't right... all I know is the units are Newtons ...

Homework Statement A solid aluminum sphere has a radius of 1.84 m and a temperature of 77.5 °C. The sphere is then completely immersed in a pool of water whose temperature is 26.7 °C. The sphere cools, while the water temperature remains nearly at 26.7 °C, because the pool is very large. The...

Ok, no luck so far on the tutor- I tried to understand and execute what you said, and it made pretty good sense, but then I got an answer that seemed way off, and sure enough, it was as the homework providor tells me it is incorrect, but here is what I did: \SigmaF[SIZE="1"]y = MA[SIZE="1"]y...

I think it is fair to call him whatever I'd like to after he literally said I was "stupid and needed to go back to high school algebra" when I asked him a question about kinetics. It is known at my school that you avoid this guy, but I am stuck with him. I'm not stupid, I have a 3.9 actually-...

A 204-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 29.6 ° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.825, and the log has an acceleration of 0.734 m/s2. Find the...

OHH! ok I get it. I don't know if I could ever do it by myself, but I get it for now. Thanks for all the help! P.S I am a girl

I'm sorry. Forget the other thread. I have never used this site before and I couldn't find where my question from three days ago went, so I started a new one last night. I thought it got rid of my old one, I guess not, but anyway read what I wrote above and tell me what you think.

ok so I solved for time in this equation: x=Vi(t)+(1/2)(a)(t)^2 but I am confused on where it came from. The equation my prof. gave me is: Xf=Xi+Vi(t)+(1/2)(a)(t)^2 so wouldn't Xi be 2400m and Xf would be the ground or 0? Or if Xi is zero, Xf would be -2400. Either way, Xf would be...

I don't understand... I don't know how to describe this angle to make sure we are talking about the same one- It is the same angle I described to you earlier. Ok there is a triangle. The hypotenuse is your line of sight from the plane to the ground. the top of the triangle is the plane. the...

I agree, that is another way I tried to do it and got the same answer, but that came up wrong too.

There is no parabola, the plane just shoots the flare down (as it is landing) and the flare has a curved path from the plane to the target, but never goes up. They want the angle between the target and the ground.

ok, so just to make sure, I solved for t and it is 13.0481. So multiply that by Vxo (that I originally calculated, right?) and you get 2711.395m So the angle is 41.514 degrees?

I don't know what you mean. Did you read what I said about the path of the flare being curved? And what quadratic are you talking about? there is only t^2 in the equation he gave me above... right?

No, actually it is landing... does that change anything? If I do what you said in the first place though I get that the angle is 2.587 degrees, which can't be right. It's my last try so I have to make sure I don't get it wrong. In the picture: The plane's nose is facing down and at it's tail...