Tetra

Really? Why not?

? What am I missing..?

Th two equation are: F net = m*a F net = m* ([v2 - v2]/t) There is also F friction = mu*F normal We know: mu = 0.10 v1 = 0 t = 8 a [down] = 9.8 an angle = 20 degrees It's possible to find the acceleration of the object going down the slope. That would be a (angled) = 9.8 sin20. But unless I'm somehow really blind and am missing out on something (probably a mass or force), then I'm not sure where to go next. My pathetic attempt of sketching it out is in my avatar.

We have insufficient information. Without a mass, we can't calculate the force of anything. And I would draw a picture (online) if I could...but it'll only let me attach URLs?

Ok, ignore the last two questions.I hadn't realized that the tension was the opposite of the force of gravity (and then calculate in necessary angles and acceleration...) So for the first question, we have the coefficient of 0.10 mu. I know we can find F net from that, but we don't have anything else. Can;t upload picture... =(

Three questions. 1) A skier has just began descending down a 20 degrees slope. Assuming that the coefficient of kinetic friction is 0.10, calculate the acceleration of the skier and the final velocity after 8 sec. We don't have enough information to solve this...or am I missing something? 2) A traffic light with a mass of 30kg is hung in equilibrium by 2 wires, one at an angle of 45 degrees, the other at an angle of 37 degrees. Find the tension in the two wires. 3) One bucket weighing 25N is hanging by a massless cord from another 25N bucket, which in turn is hanging from a third 25N bucket. All three are being lowered with an acceleration of 1.5m/s^2. Calculate the tension in each cord.

Yeah, that's what I was wondering. I was considering constructing a right triangle to visually show the thing process, then solve with trig. The hypoteneuse would be the 30m/s, and the hieght would have been the 1.5m height of the student. But we can't do that can we, because we don't necessarily know that the student is at the max height, or farther. Butif we can't then we don't really have any information to work. on. How could we find the height when we know only two vertical variables (acceleration and some height?)

So if the bar is at its highest, then it wouldn't be propelled by anything anymore, thus only gravity acts on it. (so 9.8m/s^2). And because there is no change in the position of the object at the max, then velocity would be...0m/s? As for the horzontal, a horizontal projection (component) would have 0 acceleration, while it's velocity is still moving at 30m/s. Is that right? Because even at the max height, it's still moving forward at the constant speed (assuming there is no air resistance or anything).

Just three questions that I want to clarify about the meanings for. Q1: A northern wind (blowing north to south) is blowing at 70m/s. If a jet is flying at 280m/s and the pilot wanted to fly due west, what heading should the plane take? What is the new groundspeed for the plane? Ok, so I Just want to check: this question is asking for what angle the plane should be flying at (somewhere in the NW direction) so that, with the northern wind, it would be going exactly west? And then I find the ground speed too, yes? Q2: Mr. Smith throws a bar at with a speed of 30m/s at an angle from the horizontal that gives maximum range. His student is sitting on a desk located 1.5m above ground level ready to catch the bar. a)What are the horizontal and vertical components of the acceleration vector at maximum height? What about velocity? b) What is the max. height the bar can reach? For a), I;m not sure what the difference between the acceleration and velocity would be for the components. Aren;t they the same in this question? And is it asking for them at the instantaneuos moment at maximum height? And for b), the max height would be 1.5, right, because the student is at that hieght to catch it? Or should we just assume there is no one there to stop it?

Oh hey, thanks! I set up the equations differently (had y be the height of the tower, and y0 be the velocity, since they're different for both), and got the right answer! Thanks man. And if you wouldn't mind...could you help me with this other one too? Don't worry, I won't ask for any more "From a point 70 m above the ground, an object is projected vertically upwards with a velocity of 25 m/s. How long will it take to reach the ground, assuming acceleration is 10 m/s^2." Similar to the first question, except we have a distance... and we don't know when the object starts to fall.

Well (and I think THIS is where I did it wrong) A: 10 m/s^2 V (initial) = 60 m/s I guess here I must've assumed that V (final) was 0, because then: a = (V1 - V2) / T 10 = (60 -0) / T T = g0 / 10 T = 6 Hey, should I just, I dunno, PM you or something just to make it easier? Than talking through a thread, that is...

So what could the final velocity be, then? And it's not necessary to calculate for the answer, anyways. You only need V (initial) for that equation. Are any of the other parameters (possible somehow) wrong?

I'm not sure what i'm doing wrong for the following question: "An object is dropped from a tower. If it had been thrown down at 60m/s, it would half taken have the time. What is the height of the tower? (Note that acceleration is 10m/s^2 here)" So the parameters should be: V (initial): 60m/s V (final): 0m/s A: 10 m/s^2 T: 6 s And if I used the equation D = V (initial)*T + 1/2 a(t)^2 and subbed it all in, I ended up with 540m for the height. But the answer is supposed to be 320. Does anyone know what I did wrong?

"Fling my paper"? You mean use as much space as possible? If so,then I think I get it. Thanks!

Not sure if this is in the right thread, but... We had finished a lab in which we were supposed to observe and take pictures of budding yeast cells We're now supposed to make a biological drawing. Since the yeast cells are so small, they fit across the field of view about 50 times or so, but we only need to draw one of them (budding). Are we supposed to scale them so that they would cover more space on the page, or keep them in their original size? And if we scale them, do we also have to indicate the scale? (e.g. 1cm of our drawing = 0.05cm of the actual?) I'm just wondering because, in my beginner years, we were taught to draw both an actual diagram AND a scaled one.

This is my interpretation of the method. Is it correct? If you divide the number of oxygen moles by 5 (in this case, 2.75mol) then that means that each mole of the compound would be 0.55 mol? And we multiply by 2 to get the mass of element X? Then, to find the molar mass, we divide the mass by the number of moles (2?) to get the molar mass? The the closest answer would be hydrogen...I don't think I did that right...

Hi, I'm a little new to the field of chemistry, so if this question seems a little too basic, then sorry for that. The questions is: a compound has the formula X (subscript 2) O (subscript 5) where X is the unknown element. The compund is 44% oxygen by mass. What is the identity of elemnt X? What I thought of doing was that, by converting it into grams, 44% O is 80g, but wasn't sure how to work from there (If I'm even suposed to do that.) Could you also please explain the (proper) procedure? Thanks