Physics
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OpenStudy (anonymous):
2 Questions based on KINEMATICS......HELP REQUIRED!!!!
12 years ago
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OpenStudy (anonymous):
12 years ago
OpenStudy (anonymous):
@agent0smith @Ashleyisakitty @april115 @andriod09 @BulletWithButterflyWings @e.cociuba @terenzreignz @timo86m @yrelhan4 @Phoenixuncaged @Luigi0210 @Loser66
GUYS/GALS PPPPPLLLLLLLLLLSSSSSSSSSSS HELP!!
12 years ago
OpenStudy (anonymous):
@NaCl @nincompoop
12 years ago
OpenStudy (anonymous):
:c I dont know how to do dis -.-
12 years ago
OpenStudy (anonymous):
No prob..Cheers!!....But I really need the friggin ans/method!!
12 years ago
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OpenStudy (agent0smith):
Attaching it sideways really doesn't help :P
12 years ago
OpenStudy (anonymous):
are you there?
12 years ago
OpenStudy (anonymous):
exir.ru/solutions/Kinematics.htm this could help .. Found few similar sums
12 years ago
OpenStudy (anonymous):
@agent0smith @Ashleyisakitty @april115 @andriod09 @BulletWithButterflyWings @e.cociuba @terenzreignz @timo86m @yrelhan4 @Phoenixuncaged @Luigi0210 @Loser66
12 years ago
OpenStudy (anonymous):
AC looks like diameter right ?
12 years ago
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OpenStudy (anonymous):
Ya
12 years ago
OpenStudy (anonymous):
well g along AC (call it gc) equals g
g along AD (call it gd) equals gcosa
and gb equals gcos(a/2)
agree so far ?
12 years ago
OpenStudy (anonymous):
Figured that!!
12 years ago
OpenStudy (anonymous):
now, if AC is the diameter so
AB = AC*cos(a/2)
AD = AC*cos(a)
12 years ago
OpenStudy (anonymous):
agree ?
12 years ago
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OpenStudy (anonymous):
Yup
12 years ago
OpenStudy (anonymous):
so now use
y = gt^2/2
12 years ago
OpenStudy (anonymous):
Wat answer do u get.Pls post ur work!!:)
12 years ago
OpenStudy (anonymous):
for AB we get:
AC*cos(a/2) = gcos(a/2)*(tb)^2/2
AC = g(tb)^2/2
12 years ago
OpenStudy (anonymous):
looks like the times are equal
12 years ago
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OpenStudy (anonymous):
so c,b,d are the answers
12 years ago
OpenStudy (anonymous):
yes it is
12 years ago
OpenStudy (anonymous):
On to the next one,I guess!!
12 years ago
OpenStudy (anonymous):
mmm
12 years ago
OpenStudy (anonymous):
i did something, i would like to hear though what would you do
12 years ago
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OpenStudy (anonymous):
i mean, i got one of the answers there
12 years ago
OpenStudy (anonymous):
take v=5t for 1st case!!
I will take u=v=5t.......for second case!!
12 years ago
OpenStudy (anonymous):
|dw:1376136491172:dw|
12 years ago
OpenStudy (anonymous):
Is my meth correct??
12 years ago
OpenStudy (anonymous):
vav = (v1 + v2)/2
vav = (0.25v,vsqrt(3)/4)
|vav| = v/2
12 years ago
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OpenStudy (anonymous):
that is what i did anyway
12 years ago
OpenStudy (anonymous):
How did u get v1=0.25v??
12 years ago
OpenStudy (anonymous):
no, v1 = (v,0)
12 years ago
OpenStudy (anonymous):
So wats 0.25v....Pls sho the calculations!1
12 years ago
OpenStudy (anonymous):
v1 , v2 are vectors .. look at what i draw
12 years ago
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OpenStudy (anonymous):
v1 + v2 = (v,0) + (-vcos60,vsin60) = (0.5v,vsqrt(3)/2)
so vav = (v1+v2)/2 = (0.25v,vsqrt(3)/4) <- this is a vector!!
now take its magnitude and get
v/2
12 years ago
OpenStudy (anonymous):
The drawing's not helpin...:(
CAn u do a wrtiten post??
12 years ago
OpenStudy (anonymous):
*written
12 years ago
OpenStudy (anonymous):
Hand drawn I mean!!
12 years ago
OpenStudy (anonymous):
Or draw it a bit btr with the sides!!
12 years ago
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OpenStudy (anonymous):
|dw:1376136880838:dw|
12 years ago