ok ... i am getting some of it . Let us take a base like NH^{-}_{2} . This is a very good base . But as far as i know , this does not abstract the hydrogen , but attacks the carbon center to form amides .
Homework Statement
Why do bases like OH- etc. remove \alpha-H from carbonyl compounds like ketones and aldehydes (aldol condensation) whereas amines and derivatives like NH_{3}attack the carbon centre of carbonyl groups ?
Homework Equations
none.
The Attempt at a Solution
Thought...
this is not a homework ques, but a general question..
Homework Statement
if there are two functions ... f(x) and g(x) , then what does f(x)|g(x) mean. that's it.
Homework Equations
The Attempt at a Solution
and now i am sure that the right air column plays a role because initially when the tube was in horizontal position the pressure of air columns on both sides was P . hence when tilted the volume of the right air column was increased by a small margin . this is wat it looks like :
see i think since everything is taken in terms of length of mercury column , g and density of mercury are canceled out leaving the ans in cm of mercury . that is why the answer is 75.4 cm of mercury also this is the reason why 5cm is used .
p1 and p2 are also just pressure in terms of the...
well , mass= volume * density;
now a =area of cross-section and length of mercury column=5
mass = A * 5 * density . now i am really confused . the answer evidently is even greater than 100 . ans actually is 75.4cm of mercury . ok i was wrong about the A being in the final equation , but still i...
ok , i get it , so is the eq . something like this;
45.25*A*P = (P+\frac{mgsin60}{A})*A*44.5
but it gives a linear eq . undefined in A and P . wat bout that. the ans is supposed to be 75.4cm.
one more thing if by any chance the acc is greater than g , the block tends to move upwards rather than downwards , because it actually stops when a=g . please feel free to point out errors if i made any in saying this . so as far as i know the answer can be anything but positive . but if u r...
see when the mass slides down , the diagonal component of it weight is mg sin(\theta) whereas the pseudo force component acting diagonally up is mAcos(\theta).
even then if u put a=3g , then the answer is -g\sqrt{2}.
i think this is how it should be :
the eq for banking is : tan (\theta)= \frac{v^2}{rg}
where r is the radius and g is acc due to gravity . tan theta remains constant leaving u with this
\frac{13^2}{r} = \frac{26^2}{r'}
now r' =4r
this should be the answer.
Homework Statement
A thin tube of uniform cross-sectionis sealed at both ends . when it lies horizontally,the middle 5cm length contains mercury and the 2 equal ends contain air at the same pressure P. When the tube is held at an angle 60\circ , then the lengths of the air columns above and...
Help neede in coordinate geometry!
Homework Statement
here is the question :
If the pairs of lines x^2-2pxy-y^2 and x^2-2qxy-y^2 are such that each pair bisects the angle between the other pair , then pq equals ...??the answer somehow is -1. i think it uses the concept of homogenisation (if u...