Dynamics (and Thermo)

1) We have a problem dealing with acceleration in terms of velocity (weird, I know, but it's what it is). Here's the skinny:

a = -2v ft/s²
x₀ = 2 ft
v₀ = 10 ft/s

My book shows you solving this by using dx = v*dv/a to get x, so I got for the distance the particle falls before stopping, 22ft.

It also shows using dt = dv/a to get t, which unfortunately ends up as -.5 ln(v/vo), and while solving to find the time at which v = 1 ft/s, it works fine and I get something like 1.15s, but for v=0, well, it's a natural log, and natural log of zero is [lol].

2) We have a problem with a pressure cooker, which I think is retarded to begin with. It says the pressure outside the cooker (the atmosphere) is 101KPa. The pressure inside the cooker (gage pressure) is 100KPa. This makes a net pressure difference of 1KPa, right? Idk.

You have a little piece of metal that the pressure on the inside will force up if the pressure gets too high, and the idea is to find the mass. The metal is 4mm² in area.

I did the sum of the forces = 0, which is the net pressure times the area plus the mass of the metal times gravity, and infuriatingly, my answer was off by a [couple] order[s] of magnitude. Instead of 40.8g, mine had a mass of .4077g.

1) I got the particle stops when x=3 feet

I used

-2v=dv/dt
1/-2v=dt/dv
t=-.5lnv + c
t = 0, v = 10
t=-.5ln(v) + c
0=-.5ln[(10)] + c
c=.5ln10
t= -.5lnv + .5ln10
t= -.5ln(v/10)


Bah im too rusty

2)I dont know... I cant take the pressure!!!![/pun]

Figure it out yourself.

.408kg is 408g, son. You multiplied by 100 not 1000. And I already converted to grams anyways. And you converted 4mm² to 4meters by milimeters. You have to divide by another 1000 to get to meters square.

t = -1/2 ln v/v0 <- that's what I got and mentioned in the post.

1.15129 sec <- That's the time when v=1. It doesn't make sense that it should be 1ft/s and 0ft/s at the same time.

You might want to go back to physics?

I get snappy when people are condescending, that's all.

The atmospheric pressure was 101kpa, the gage pressure was 100kpa. Gage is pushing up, atmosphere is pushing down, their sum is 1kpa, isn't it? Otherwise it would be 201 which seems even less likely.

Quote:

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Originally Posted by rubah

I get snappy when people are condescending, that's all.

The atmospheric pressure was 101kpa, the gage pressure was 100kpa. Gage is pushing up, atmosphere is pushing down, their sum is 1kpa, isn't it? Otherwise it would be 201 which seems even less likely.

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Quote:

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Originally Posted by wikipedia

For gases, pressure is sometimes measured not as an absolute pressure, but relative to atmospheric pressure; such measurements are called gauge pressure (also sometimes spelled gage pressure). An example of this is the air pressure in an automobile tire, which might be said to be "220 kPa/32psi", but is actually 220 kPa/32 psi above atmospheric pressure. Since atmospheric pressure at sea level is about 100 kPa/14.7 psi, the absolute pressure in the tire is therefore about 320 kPa/46.7 psi.

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Why don't you give that a good read.

And I wasn't even being condescending, just happy to finally be the first one to help someone. I just read something wrong(you get like that when you're in a hurry but still want to help someone), forgive me for being human.

But if it's this bothersome to you, I will refrain from it in the future, I don't need the grief.

Threads get confusing when people delete posts. 0_o

Yes, you were right about the pressure thing. I was asking for clarification on that :p I was mad because you were bragging about how much more physics and engineering you'd studied, because it had nothing at all to do with the material at hand and just made me feel even stupider for having trouble with something this easy. I'm sorry I was cranky when I woke up this morning and that you were the first good target I came upon :p I appreciated the help, just not all the extra things you said.

Anyways, what he deleted about the pressure ended up being that the absolute pressure of the cooker cancels out the atmospheric pressure, leaving only the gage pressure behind, which cancels the weight of the petcock. So you set the gage pressure (net pressure) equal to mg/A (mass gravity, Area) and solve.

The problem with the accelerations apparently I had right, even if I didn't believe it. I talked to my professor and he said it was acceptable to have an infinite amount of time to come to rest xD weird!

Quote:

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Originally Posted by Aerith's Knight

And I wasn't even being condescending, just happy to finally be the first one to help someone. I just read something wrong(you get like that when you're in a hurry but still want to help someone), forgive me for being human.

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I think she was referring to the general tone of the post, especially the statement that was something along the lines of "this is High School Physics to me." Whether or not it is or isn't to you does not guarantee that the person you're helping is in the same place. When you combine a statement such as the given with getting basic "High School" things such as unit conversions wrong, it definitely does come off as something that is deserved.

Physics: serious business.

I believe that the pressure difference would be 100 kPa, if it helps, not 1 kPa, which is why your original answer was off by two decimals. The atmosphere outside of the cooker is the reference point; gage pressure is the difference. The rest of problem number two seems like it is just finding the right conversions and canceling out units.

cba to do the first one.

WAY TO NINJA MY POST RUBAH