Liq sucks at physics. Don't know why anyone would ever want his help.
Spoiler (highlight to read):
<3
Spoiler (highlight to read):
<3
how did you get v2x = 20.5(cos32.7))liquidat0r wrote:
Question 2
Hopefully you can read/decipher my writing/working.
Would type it out and make it look nicer and easier to follow but I don't have time.
http://i44.tinypic.com/2evyf75.jpg
The v's with subscripts are meant to be like these:
v1y = Initial velocity in y direction.
v1x = Initial velocity in x direction.
v2y = Final velocity in y direction.
v2x = Final velocity in x direction.
(Same goes for the Forces (F) and accelerations (a).
edit: never mind it's an 28.5, it looked like a zero, lol
Last edited by TheunforgivenII (2009-01-27 09:14:54)
Does this help for question 2?
What are you actually supposed to find? Note that my theta probably isn't the answer, it'll be 90 minus theta, or something ... probably.
What are you actually supposed to find? Note that my theta probably isn't the answer, it'll be 90 minus theta, or something ... probably.
really is nice writing...
Small hourglass island
Always raining and foggy
Use an umbrella
Always raining and foggy
Use an umbrella
(a) In what direction should you head? (in degrees)liquidat0r wrote:
Does this help for question 2?
What are you actually supposed to find? Note that my theta probably isn't the answer, it'll be 90 minus theta, or something ... probably.
http://i43.tinypic.com/107masi.jpg
(b) How long will it take you to cross the river? in (seconds)
(a)
Well I've given you an angle. Presumably you either need a heading or to say "X degrees from the positive X direction", or something.
(b)
You have a distance across the river and I think you should use the y component of the velocity (like I used the x component before). You can work out the time taken using s = vt
Well I've given you an angle. Presumably you either need a heading or to say "X degrees from the positive X direction", or something.
(b)
You have a distance across the river and I think you should use the y component of the velocity (like I used the x component before). You can work out the time taken using s = vt
I'm having some problems on figuring out the direction. I'm used tan-1(fy/fx) which would be tan-1(15.39/23.98) right?liquidat0r wrote:
Question 2
Hopefully you can read/decipher my writing/working.
Would type it out and make it look nicer and easier to follow but I don't have time.
http://i44.tinypic.com/2evyf75.jpg
The v's with subscripts are meant to be like these:
v1y = Initial velocity in y direction.
v1x = Initial velocity in x direction.
v2y = Final velocity in y direction.
v2x = Final velocity in x direction.
(Same goes for the Forces (F) and accelerations (a).
edit: never mind I figured it out, it's tan-1(15.39/6.58)
Last edited by TheunforgivenII (2009-01-27 10:10:52)
okay well I found what direction to head which I did by cos-1(0.5/1.6) = 71.790043. Then I subtracted 71.790043 by 90 degrees and got 18.20, which is the correct answer. But I still can't figure out How long will it take you to cross the river. I started to use the width of the river/vy where vy=vrow(sin[x]), x is the angle with horizontal, but I can't seem to plug the right numbers in.liquidat0r wrote:
Does this help for question 2?
What are you actually supposed to find? Note that my theta probably isn't the answer, it'll be 90 minus theta, or something ... probably.
http://i43.tinypic.com/107masi.jpg
Looks like it'd be a hell of a lot simpler if you used unit vectors 1st then worked it out from there.TheunforgivenII wrote:
okay well I found what direction to head which I did by cos-1(0.5/1.6) = 71.790043. Then I subtracted 71.790043 by 90 degrees and got 18.20, which is the correct answer. But I still can't figure out How long will it take you to cross the river. I started to use the width of the river/vy where vy=vrow(sin[x]), x is the angle with horizontal, but I can't seem to plug the right numbers in.liquidat0r wrote:
Does this help for question 2?
What are you actually supposed to find? Note that my theta probably isn't the answer, it'll be 90 minus theta, or something ... probably.
http://i43.tinypic.com/107masi.jpg
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
I already got the answer...all I had to do was cos(18.20)(1.6) = 1.52 and then took 50/1.52 and got 32.89presidentsheep wrote:
Looks like it'd be a hell of a lot simpler if you used unit vectors 1st then worked it out from there.TheunforgivenII wrote:
okay well I found what direction to head which I did by cos-1(0.5/1.6) = 71.790043. Then I subtracted 71.790043 by 90 degrees and got 18.20, which is the correct answer. But I still can't figure out How long will it take you to cross the river. I started to use the width of the river/vy where vy=vrow(sin[x]), x is the angle with horizontal, but I can't seem to plug the right numbers in.liquidat0r wrote:
Does this help for question 2?
What are you actually supposed to find? Note that my theta probably isn't the answer, it'll be 90 minus theta, or something ... probably.
http://i43.tinypic.com/107masi.jpg
indeed lulz.FatherTed wrote:
really is nice writing...
1) A 2.50 kg object is moving along the x axis at 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (plus or minus). F1 = 20 j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2.
I got one last problem but I don't know where to start
I got one last problem but I don't know where to start
guess everybody is tired of working on physics, especially me, but still need the help, lolTheunforgivenII wrote:
1) A 2.50 kg object is moving along the x axis at 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (plus or minus). F1 = 20 j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2.
I got one last problem but I don't know where to start
I got a minute, let me get a pen and paper im bored
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
Right...TheunforgivenII wrote:
1) A 2.50 kg object is moving along the x axis at 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (plus or minus). F1 = 20 j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2.
I got one last problem but I don't know where to start
First you can ignore the x values as nothings happening along that axis (proven by 1.6m/s * 3s = 4.80m it's finish co-ordinates)
y axis time.
Next...
finding the acceleration given by F1 by using F=ma
20N=2.5a
so a=8ms-2
Its co-ordinates found on the y axis due to this force can now be found:
s=ut+1/2at2
as the initial velocity is 0 the "ut" part can be ignored leaving: s=1/2at[sup2[/sup]
plugging in the values for the force f1
s=1/2 (8) * (3)2
s=4 * 9
s=36m
This shows that the other force f2 has to be in the other direction as the value 36 is greater than the 10.8 its at at that point.
Soooo finding out the resultant displacement it has to produce on the same block would be done by
10.8 + ? = 36
? = 25.2
Again using s=1/2at2 (ignoring ut as u=0)
25.2=1/2 a 32
25.2=4.5a
a= 5.6ms-2
and using F=ma
F=3 * 5.6
F=14N
F2 = 14N in the opposite direction to F1
I think, though I have a feeling I went wrong somewhere or left something out cos im tiiiiiiiired... xD
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
I'm too tired at the mo to type out the answer, but here's what you need to do - take the force (F1) given and work out where the object would end up, after 3 seconds, if only that force had been applied to it - subtract the given y from the y part of position just calculated - from that you can than work out the force required to move the object that distance in 3s.TheunforgivenII wrote:
guess everybody is tired of working on physics, especially me, but still need the help, lolTheunforgivenII wrote:
1) A 2.50 kg object is moving along the x axis at 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (plus or minus). F1 = 20 j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2.
I got one last problem but I don't know where to start
And that's your answer for F2
i'll look it over and check every step. Thanks for the help!!presidentsheep wrote:
Right...TheunforgivenII wrote:
1) A 2.50 kg object is moving along the x axis at 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (plus or minus). F1 = 20 j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2.
I got one last problem but I don't know where to start
First you can ignore the x values as nothings happening along that axis (proven by 1.6m/s * 3s = 4.80m it's finish co-ordinates)
y axis time.
Next...
finding the acceleration given by F1 by using F=ma
20N=2.5a
so a=8ms-2
Its co-ordinates found on the y axis due to this force can now be found:
s=ut+1/2at2
as the initial velocity is 0 the "ut" part can be ignored leaving: s=1/2at[sup2[/sup]
plugging in the values for the force f1
s=1/2 (8) * (3)2
s=4 * 9
s=36m
This shows that the other force f2 has to be in the other direction as the value 36 is greater than the 10.8 its at at that point.
Soooo finding out the resultant displacement it has to produce on the same block would be done by
10.8 + ? = 36
? = 25.2
Again using s=1/2at2 (ignoring ut as u=0)
25.2=1/2 a 32
25.2=4.5a
a= 5.6ms-2
and using F=ma
F=3 * 5.6
F=14N
F2 = 14N in the opposite direction to F1
I think, though I have a feeling I went wrong somewhere or left something out cos im tiiiiiiiired... xD
No worriesTheunforgivenII wrote:
i'll look it over and check every step. Thanks for the help!!presidentsheep wrote:
-Physics shizz-TheunforgivenII wrote:
-Physics Question-
Last edited by presidentsheep (2009-01-27 15:48:16)
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
your answer checks out! It's correct!!presidentsheep wrote:
No worriesTheunforgivenII wrote:
i'll look it over and check every step. Thanks for the help!!presidentsheep wrote:
-Physics shizz-
I assume that is sarcasm.FatherTed wrote:
really is nice writing...
In which case, fuck you. I have no need to write neatly. If I did, I would.
No, honestly, mine is 10x worseliquidat0r wrote:
I assume that is sarcasm.FatherTed wrote:
really is nice writing...
In which case, fuck you. I have no need to write neatly. If I did, I would.
though they do say the geniuses have terrible writing!
Small hourglass island
Always raining and foggy
Use an umbrella
Always raining and foggy
Use an umbrella
Your also bad in bed (warman said so anyway).FatherTed wrote:
though they do say the geniuses have terrible writing!
nx=sinx
Dividing through by n
x=six
Dividing through by n
x=six
Fuck Israel
the rate of change in y=r^3/3...Dilbert_X wrote:
nx=sinx
Dividing through by n
x=six
Any more? Im bored
I'd type my pc specs out all fancy again but teh mods would remove it. Again.