misc. off topic questions

[User22]

Originally Posted by Jalen

For example, one question is "If a moving cart makes a sticky/inelastic collision with a stationary cart of equal mass, by what factor will the cart's velocity change?"

Although my teacher gave the class this formula to "help find the factor, if you can't do it mentally"

MiVi + MtVt = MiVi + MtVt' ('=prime)

For pretty much all of them I ended up with Vt' = .16

Now I know that 1/6 is .167,

but I don't understand how I translate that to "by what factor"

Let's say your initial velocity (Vi) was 1/36 m/s or whatever units; and your Vt' = 1/6.

Well....1/36 times 6 equals 1/6

So that means it increased by a factor of 6. A factor of something means how many times you had to multiply it. So the factor in this case is 6....

AP Physics has not taken over my life in any way

Can't wait for electromagnetism We're on thermodynamics right now :/

Edit:

From now on this will be a non-Aaronius31 thread. I'll give you guys somewhat of an enjoyment by being in this thread and not having to worry about me posting....

[user472884]

on the possibility that some Nigerian hacker hacked your computer and prevented you from seeing your visitor message, here it is reposted.


I don't really understand the formula, but I at least understand the question is just asking about the cart that was initially moving.

The initial rate of the moving cart was .505m/s, and the final was .301m/s

and that would be .596....

NO UNDERSTAND

and if it means anything, the moving cart's velocity was 1030.0g while the stationary cart's was 526.2g

Originally Posted by Jalen

Call me retarded, but I don't understand what it means when somebody says something like "it changed by a factor of 4"

and this whole physics lab is pretty much that

For example, one question is "If a moving cart makes a sticky/inelastic collision with a stationary cart of equal mass, by what factor will the cart's velocity change?"

Although my teacher gave the class this formula to "help find the factor, if you can't do it mentally"

MiVi + MtVt = MiVi + MtVt' ('=prime)

For pretty much all of them I ended up with Vt' = .16

Now I know that 1/6 is .167,

but I don't understand how I translate that to "by what factor"

Factor means the quantity to be multiplied by input to get the output. In this case, its jus ratio of final velocity(output) to initial velocity(input).

P.S. If you guys want then you can create a separate thread for homework/school problems. That would probably help everyone and will you give you another reason to visit AAm.

[user472884]

I don't think that questions are that numerous, but I'm sure it wouldn't hurt

[Azhiri]

Yeah, make a thread like that, because this Algebra exam review is kicking my butt.

and it's worth 300 points on my final average for this semester and I NEEEEEEEEEED that grade desperately or I'm staring down the barrel of a D-

HELP

[user472884]

done, for the lulz

Edit:

Oh lawdy I as me quite a decision to make.

Tomorrow, I can either stay home, sleep in, procrastinate, perhaps get some Assassins Creed in....

or I can go paintballing, for which I haven't gone in quite a while, and Zach's older sister will be going and I have permission (from Zach) to "shoot her square in the tit"

and it's very tempting to shoot a Foster tit.

but then again it's going to be like... 20 tomorrow

Edit:

- To gun people (*cough*Jung*cough*), should I attempt to make (yes, I have the means) a 4 inch rifled .22 barrel, or would it not even matter and I should save the time and effort and make it smoothbore?

I'm not looking to do shit like "in the game" (that's what the blade is for), but I'd like to be able to have a decent chance of hitting say, an apple at ten yards (assuming that I somehow aimed correctly)

[jung_adore_ALIZEE]

Originally Posted by Jalen

- To gun people (*cough*Jung*cough*), should I attempt to make (yes, I have the means) a 4 inch rifled .22 barrel, or would it not even matter and I should save the time and effort and make it smoothbore?

I'm not looking to do shit like "in the game" (that's what the blade is for), but I'd like to be able to have a decent chance of hitting say, an apple at ten yards (assuming that I somehow aimed correctly)

I'd say do a rifled barrel, somewhere in a 1 in 16 to 1 in 12 rate of twist. My 2 inch bobcat will hit a lemonade jug about 50% of the time at 25. With any modern firearm there is no reason you shouldn't have a rifled barrel, even your black powders firing round lead balls are rifled granted at a rather slow rate. Local laws may restrict making of your own firearm however this is typically set in regards to class 3 firearms.

Regards,

Jung

[user472884]

Them cops ain't got to know about it, now do they?

Now, what kind of steel should I mill it out of, and is there anything special I need to do anything to it so it doesn't blow up in my hand mess up my blade (oh yeah, and that thing about shrapnel in my wrist veins and stuff)

Or am I just really, really overestimating the power of the .22?

[AlizéeInspired]

Nnnope. F*** THA POLICE!

[jung_adore_ALIZEE]

Originally Posted by Jalen

Them cops ain't got to know about it, now do they?

Now, what kind of steel should I mill it out of, and is there anything special I need to do anything to it so it doesn't blow up in my hand mess up my blade (oh yeah, and that thing about shrapnel in my wrist veins and stuff)

Or am I just really, really overestimating the power of the .22?

4140 or 4150 are common for barrels, I'd assume it would have to be heat treated after milling however, not sure I'm not a steel worker. I'd suggest just picking up a barrel for a Buckmark (or something else) and build around that. Other than that I'd say make sure the bore size is proper and you have good lock up, don't want an out of battery firing afterall.

The risk with .22 lr is minimal but it is still there. As it is the .22 lr is typically in the 24000 psi range and you can take and compare that to the standard .45 ACP which 21000 psi, and the +P variant being 23000 psi. That being said yes a .22 lr can produce a kaboom however it is very unlikely.

Regards,

Jung