Tetra

Kw is the equilibrium constant of water...but I'm not sure what that means. As for adding OH, it would react with the H to form water molecules?

We're currently on the Equilibrium unit, and I'm having problems with some questions. 1) Given a 0.015 M solution of HOCN with a pH of 2.4, find the ionization constant, Ka. I know that ka = ([H+] [OCN])/ [HOCN], and i can calculate for the concentration of H+ and HOCN. But how do we find the concentration of OCN>? 2) Given a 0.015M solution of an unknown base with pH10, find: a) The concentrations of hydroxide and hydronium ion b) Kb I think for part a) I use the -log[H+] = 10 to find the concentration of H+. But then what about hydroxide? And as for b), Kb = Kw / Ka, but since we don't know Ka, how do we solve for this? Sorry if it's a bit confusing, but I'm also confused. I really don't understand this unit.

How about in theory? Your method was to physically add 100g of solute, then enough water to make it 1L. But what if it's just through calculations?

This is a really weird question...in my opinion... We are told to bake a cake, and we have to use a sulfur solution (sulfur + water ONLY). We must add 100g / L of the solution. (Yes, it's grams, not ml.) How much of the solution should we add to get that amount? From what I know, concentration = mol / L. We know concentration, and we have to find the volume. What I'm not sure is how to calculate for the mol (we're not given a mass) and how to convert the g to ml? I'm not sure if mass is necessary. Or, if we need one, we use a number found in society and indicate why we use it? I'm not sure if I even approached this question correctly...if anyone could help, that'd be great!

I'm not certain. it would be most preferable if they let me put a light in. I'd say it's a 50/50 chance I'm allowed to put a light in there.

I have to design a lab with an organism is a pollutant. I need 5 increments and 3 responding variables. There are three days to do the lab, with 1 hour or so a day to run experiments. For the rest of the time the organism has to live in a closed shoebox. I was thinking of using some type of water plant, like algae, since water pollutants have faster effects than airborne or soil. For the pollutant, maybe cadmium? Or Ni, or Pb. I know those are major sources of water pollutants, though I'm not sure if they'll have much effect Does anyone have ideas of what increments can be used and what can be observed? I can only think of growth, and I;m not even sure if that will work since the organism is hardly ever exposed to any light. If anyone can help, I'd be really grateful.

Hi, I'm a high school student who has to run a self-designed lab. I'm supposed to see the affects of a carcinogen on a living organism. In this lab, there has to be 5 increments/ manipulated variables (cannot be time) and three responding variables. It's also preferable that it doesn't take too long to do, maybe MAX 1 + 1/2 weeks. However, I'm having trouble thinking of ideas. 1) The carcinogen : What can I use that is (relatively) safe and accessible? I do not have access to many chemicals (unless i make it myself), and even then I'm afraid that too much exposure can lead to some health problems. I considered radiation, but I'll need strong ones, like gamma or UV to make much of a difference, so I'm not sure about that either. 2) The organism : Well, I can't take really expose a human being or animals to something that can cause cancer (unless I find some small things, like ants). I was considering plants or fungus, but that might take too long, and it'll have to depend on the lab. If anyone could help, I'd be really grateful, thanks.

Um...I muse be stupid or something... So say I used that equation v2^2 = v1^2 + 2(a)(d) and I find d. Then I find the horizontal velocity.Then i can find time, which is the amount of time it takes for the bird to drop. But since that's a vertical component, I'm supposed to use the vertical velocity to calculate for time. And, since time should be the same for both horizontal and vertical components, then I can use the horizontal component and time to find distance...is that it? I thought there was another way of doing it...I guess I must have imagined it....

So...what does it have to do with conserving momentum? And for the height, we just use one of the kinematic equations?

Don't really get what to do... A hunter shoots a 500g arrow at a 2kg bird perched on a tall tree growing on a flat ground. The arrow is launched from ground level with a speed of 40m/s at an angle of 30 degrees above the horizon. It is traveling horizontally when it strikes and embeds into the bird. How far from the base of the tree do the bird and arrow land? So I found that (using the law of conservation of momentum) the momentum of the bird + arrow is 20kgm/s, and the velocity of the arrow + bird is 20m/s. And I know we can form a triangle thing out of the velocities or momentums but...I can't seems to find distance! Which is the thing we're looking for! Help?

To Cap'n Refsmmat: Ok, I see what I did wrong now! It was a silly mistake, I drew the triangle but didn't realize that it was the hypotenuse that I was supposed to be looking for. So, for the tension in the wire, it's supposed to be force of 'gravity / sin40'. Similarly, the tension in the rod is the horizontal component of the tension in the wire, so that formula is 'Force of tension cos49'. Thanks for all your help, today and with other stuff! To ewmon: Hmmmmm I think I get your gut feeling thing. It's like visualizing the actual sin and cos waves, where it goes up and down at certain degrees. So that way we kind of get an idea of the magnitude of the actual value we're getting (small, huge, and the size relative to other objects in the situation.) (P.S. And I didn't realize before that compression meant and outwards force..thanks for clearing that up!

Late reply, but... To cypress Oh, your right...I forgot that 82 was a mass. So, if I found what the force of gravity is, then I can form a right-angle triangle. Knowing that the angle is 40 degrees, and the opposite is 803.6 (the force of gravity) I can use 803.6sin40 to find the hypotenuse, or the tension force. (I would upload an image but...it needs to be on a URL?) And for the rod, since we know (or calculated) the tension force, then we can draw a right triangle again and solve for that horizontal force...right? But I don't get it....if the object is in equilibrium, shouldn't the forces pulling on it equal 0 in the end? Like, if you break it into horizontal/vertical components, it's not 0...or then would that mean it's not right? To ewmon: Oh, I know how to formulate trig functions (at least..I'm pretty sure I did it right). But could you explain this "gut feeling" thing? I can draw everything out, but I don't get the part about "developing word answers" and whatnot. And this is supposedly physics...the dynamics unit in physics.

A sign for a restaurant has a mass of 82 kg. It is held out from the wall by a light horizontal steel rid which supports no weight and a wire at 40 degrees to the horizontal. (This will form a right triangle). Find the tension in the wire and the compressions in the steel wall. For the tension in the string, I know that once we find the vertical force of the sign (803.6 N [down]), then we can use 82sin40 to find the tension in the string. But what about the wire? Is the force just the force of the object itself (so 803.6N [down])? Also, in a question where an object is hanging from the center of a rod or string or something, will the tension in either side of the string be identical? Or is there some way for them to be different?

We're supposed to calculate average acceleration: So we're given a situation of a plane traveling some distance with four different velocities. We were taught that, to draw a velocity vector diagram for it (NOT a displacement one), we're supposed to use draw it using the first and last given velocities. Then, say we know an angle for the triangle formed, we can use cosine law to find the resultant displacement. My question is, since acceleration is a = Δv/Δt, then why do we need to find the resultant displacement, why not just subtract v2 from v1 to find the change in velocity? Why use the displacement?

This is just me musing: I think the method could only be used for problems with uniform motion, because if it weren't (a straight line), then we can't simply solve for the slope by using a = v/t. In this case, if averages are used for v and t, we can find average acceleration (which is just acceleration, since this is a straight slope) So, in this case, if we use the averages for both v and t, we can solve for a (which is 66/ 5.5). And, since the acceleration is uniform the whole way, the method can be used backwards to solve for distance in the 7th second. So a = v / t 12 = v/ 6.5 = 78 And v = d/ t 78 = d/ 1 d = 78 ...am I right? Or am I just using a confusing method..?

That's true... Oh I just thought of something! Not sure if it's right, but... If d/ t is average velocity, then, if say we were to draw it on a v/ t graph, the time of 5.5 should have velocity of 66 (because 5.5 if between 5 and 6, then the center point should have the average velocity?) So from that we can calculate acceleration. But I'm not sure about the next part of the question, which is to find how much displacement was covered in the sixth second. I was thinking we could us the equation: d = vt + 1/2 (a) (t)^2 But I'm not sure what to put in for time. Should it be one second, because one second elapsed in the 6th second?

Just a (simple) problem that I couln't solve. A motorcycle starts from rest and, moving with uniform accelerations, moves 66m in the sixth second (from time equals 5.0s to time equals 6.0s). What is its acceleration, and what distance does it travel in the 7th second? I'm not really sure how to do this. Why can't we do the following? So to find acceleration, the equation is a = Δv/ t, or (v2 - v1) / t. And v2 = d/t. So v2 = 66 /1 = 66 So, to find acceleration: a = (v2 - v1)/t = (66 - 0) / 6 = 11 But that's not right! And a friend told me it was supposed to be 66 / 5.5, but he doesn't explain why either! So I don't get it. Help?

Ohh I see. Yeah, I thought it was something really complicated. I didn't think about dampening before, but now that I do, it makes sense. Thanks.

Yes, but why does it keep on echoing? Say if I were tap a metal pot on the edge, or the rim of a (plastic( bucket, the echo woulnd't last as long. Is it reverberation? Or the shape of the object? Maybe using a pot was a bad example. How about a cymbal? I'm guessing the concepts behind them are similar.

I don't understand why, when you tap a metal pot lid, it has a really high pitch as well as resonates. I understand (or I think) that if you hit something, it will vibrate, and since a pot lid is a circular membrane, it will cause resonance, but I'm not sure why. And is the pitch just because of the material itself? Or is there some other reason? Does it create longitudinal waves or transverse? So confused. @_@

So stress (is it the same as pressure) is the amount of force applied over an area, and the strain is the amount of stress an object can withstand? Are there any ways to calculate strain? I've looked around, but it's been referred to "deformation" a lot, though they don't really explain what it is except the basics and some applications.

This isn't exactly a homework question, so it might not be in the right thread, but... We've been learning a lot abut forces and such, as well as momentum and impulse.There were two things I was wondering about. 1) How do you calculate the amount of foce needed to break something? For example, a board. Do we just use F = ma to find out how much force the board has? And if so, does that mean if we strike the board with that same forcve, or more, then it would break? 2) How does force injure a human body? I've never really thought of it before, but the more I think about it, the less I understand. Is it that the bodt is resisting a change in acceleration, and thus it is injured? Or that it's due to force that's just breaking some of the cells or such in the body?

All right, I understand now. Thank you!

Well, that DOES make it easier for the eyes =P Just one thing I'm not sure about the calculation. When you were grouping it at: =-m*sin(20)g+µm*sin(20)g =mg*sin(20)(-1+µ) Wher did the -1 come from? Wouldn't it remain poistive? And how is (m*sin(20)g) the F grav? Wouldn't that just be the fricitonal force?

O_O OK....I'll have to work through that,,,might take awhile... Haha...thanks..?