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IS THIS A DOUBT OR A CHALLENGE
NE WAY
1) 8*rt2* a
2) I THINK THIS QUESTION IS INCOMPLETE
3) TRIANGLE TAPQ WITH FOUR VERTICES ???? I DINT GET U
4) a)
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1) Well I Didnt get the ans man I think theres more to the que
2) RyuAmakusa
THE CASE IS IS IMPOSSIBLE becos f(0) > 0 and graph cant be continuous with roots btw 1,-1
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A B C if l is slant length
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Nuksi
The ans which u gave is definitely wrong bcos it eventually bcoms the eq of 2 imaginary lines
Yes it is a combination of rot and shift of axes
eq is 2X^2 - 3Y^2 = 1
Formula for rot of axes is
X= (xcosO - ysinO)
Y= (xsinO + ycosO)
If eq of line is sometin else apply the shift in axes as X+ h and Y +k
h = -3 k = -2 ( lines intersect at 2,3 )
tanO = -1/2 ( slope of transverse)
sinO = -1/rt5 cosO = 2/rt5 ( O in 4th quad)
substitute and get
2 ( 2X + Y - 3)^2 - 3( 2Y - X - 2)^2 = 5
good question
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1 for first eq it is in the form of root of a sq
sq my ans and see ull get the exp
2 In the second one sorry I made a mistake
THE ans is 3
Let the exp be X
Cube the exp
X^3 = 12 + 5 X ( try it urself and see)
hence X=3
If u dont get it reply I will ans
3 again wat is - -
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AT THE MIDPOINT OF DOOR EDGE
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SO THATS THE MISTAKE
IT IS 1.5 FT
THEN THE ANS WILL BE 43N
TORQUE ACTS ON A DIRECTION PERPENDICULAR TO PLANE OF DOOR
MAN THESE ARE BASICS
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THE method involved is area of med.triangle = 1/3 ar of triangle
Solve the det to get ans
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learner the error u made is finding V as .19 Its .1973684211
put V as UF/U-F to avoid approx
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Sorry adi the formula you used is applicable only if d is negligible when compared to focal length
The correct ans are .308 , .148
-.058, .0367
Check it for urself !!!!
cheers
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Dont go by the answer in the book . It may be wrong or you may have typed 6 ft instead of 6 m or 1.5m inplace of 1.5 ft
Force is split into 2 components bcos horizontal comp balances torque and vertical comp balances weight. In the fig which ramyani gave the horizontal comp must be opp directed to maintain hor eqm
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both assertion and reason are incorrect
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A simpler solution is
Tangent is Xcos?/2 + Ysin?/4 = 1
It passes thru centre of circle 2,4
Therefore cos? + sin? =1
? = pi/2 or 0 is general sol
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Ramyani though u gave ans as 43 its wrong bcos u ignored the feet- metre conversion
on balancing torque about one corner on hindge side
(2*fcost + 2*fcost)* .3048 = 80*0.75
fcost = 49.2 --------1
on balancing forces
fsint + fsint =80
fsint =40 ---------2
using 1 and 2
f=root(40^2 + 49.2^2)
=63.4 N
Or maybe Ramyani gave correct ans to a wrong question
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Its not infinite
ITS
(f/f-1)^2 *((M+m )/ M)* V
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