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Why this question
anyways the
answer is
M^0 L^1 T^ -1
so it becomes LT^ -1
make sure X^1 means x square.
do ask if u need the explanation.
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Thanks for sending in ur doubt
see u require 50% of pure calcium carbonate
MW=100
Wt of calcium carbonate required = 10g
so
wt of pure ca. carbonate req. = 20% of 10g ie 5g
Volume of HCl has been asked here to decompose CaCO3
Lets take the volume as V
so
now as per the given eaction above
Meq of I mole of HCl = Meq of 2 Moles of CaCO3(pure)
w/MW= 2 (pure) W/MW (2W as there are 2 moles of caco3
here we take w=Density * volume
1.825 * V / 36.5 = 2 * (5 / 100 )
1.825 * V = 71 * 1/20
V = 3.65 / 1.825
V = 2mL
So the answer is option D
I would be happy if I helped u in some way or the other
plZ rate me for my efforts
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hey Ramyani I am 100% sure that the question is wrong if the answer u r saying is given that the block will move.
The force is acting in the direction parallel to the wall of 15N and that too if is acted in the direction opp to the force of 40N then there is no chance of the block to move as in
either of the cases the limiting Friction force is larger than the force that is applied II or perpendiculer to the wall.
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hey mudita Now I am dam sur ethat the number of solutions is "2"
If u draw the graph of [x]=2x+3
then u will get one solution at x=-3
and if u draw [x]=2x-3
then u will get only one solution at x=3
Thus there are to solution taking both the cases.
plZ Rate me if I could solve ur problem.
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Thanks for the rating srujana and I more happy that u have understood my concept that i wanted to explain .
Please ask me anyother doubts u have.
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hello srujana are u there I hope u understood it now
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see u really need not get the vector law in the picture and confuse ur self .
OK then centripetal force is nothig but any external force like gravitation,tension,,friction etc. The concept of centripetal force is applied to be negative as per ur question above when the magnitude is mv^2/r , but in case of non uniform circular motion the centripetal force is negative in magnitude .
explanation of neg force::::::::::
if speed of a particle in a circle is not constant then the acc has both radial and tangential acceleration .
So when the accn is radially inward that is towards the centre then then it is negative incase of a car moving in circular path and then drifting in the direction opposite to the direction of the centripetal force . I hope that this gave u some idea of the cases of when centripetal is negative and when is it positive.
please rate my efforts.
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no not at all
I mean to say that
if
you get f as negative then it is centripetal that follows the vector law if you know it . and if positive then it is outward as same as when we say that if a is negative in case of freely falling body then the body is going upwards and if it is positive we say that the motion of the body is downwards in the direction of the gravitational force. understood.
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Another point apart from the above two is that
if the force that is
F is negativa then the force acts in inwards direction and acts in outward direction in case F is positive
this can be explained as F=ma
and here a will be negative if it is inward as incase of a pure rolling body the V acts in the outward direction adn thus the A becomes opp of that of V and thus A becomes negative and becomes radially inward (CENTRIPETAL)
&
When the A is radially outward that is in the direction tangential to the V as per rotational Dyanamics than A is radially outward and thus is positive (CENTRIFUGAL)
PLEASE RATE MY EFFORTS IF THEY HAVE CLEARED UR PROblems.
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I would like to ask BIPIN that how did u get the RHS in the above question .
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see it is as simple as a piece of cake
suppose u have 2 functions given above are IxI and IyI
draw the graphs of both the equations and then for finding the graph of f(x)= max { IxI,IyI } find the graph which has the highest value for a every particular interval of X-axis
and
then for minimum u have to select the graphs that have the least value in a particular interval of X that must be less than 2
or
find the max and min values b/w minus infinity to 2.
please rate me if u think I helped u in some way or the other
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See udita I have tried the Question even by the graph mathod I think the answer is definitely 2 solutions in either case
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Atlast in the above problem it will be V component along the vertical as "u" and resolving along the line of force that is along the line where tension acts we get finally as u=vcos(theta) and in the end it becomes as V=U/COS (THETA)
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Catch_arnie is perfectly right but u can also do this by the method of resolving Tensions with the velocity and by taking the "v" components along AM and BM
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Dude u gonna try the main BIBLE OF PHYSICS
""""""""" D.C.PANDEY"""""""""""""""".
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No offenses dude
just got off the track
my mistake
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I do not agree with Catch_arnie
There will be two solutions but not that are given above
there will be two cases
1)[x]=2x+3
2)[x]=2x-3
We know that [x] can be written as (-1-x) so putting the value of [x] in both the above cases we get two solutions
x=-4/3 or x=2/3.
please correct me if I am wrong
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If  ,  are the roots of the equation ax 2+bx+c=0 then [ x] [ ](ax 2+bx+c+1) 2/x-is = a) 2a ( - ) b)2 ln I a( - ) I c)e 2a(--d)e a^2 I-I Answer is option (C)
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dude first you know that sinx/x is nothing but 1. after you take "x" common you will get
tan2x - 2tanx/4sin^3 x and then apply formula for tan2x
that is
2tanx/1-tan^2 x
then
u will get 2tanx commom
with tan^2x left common inside
atlast
it will be
2tanx * tan^2x/4sin^3 x
place value of tanx and then u will get
1 / 2.cosx.cos^2 x
ans is 1/2 if u place the value of x=o in the given equation
you must have solved it.
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