114 - Kinematics Questions Answers
Q A VECTOR A POINTS VERTICALLY UPWARD & B PTS. TOWARDS NORTH. VEC. PRODUCT ACROSS B IS ALONG WHICH DIRECTION?
Q LET < B/W TWO NON ZERO VECTORS A & B BE 120* & ITS RESULTANT BE C THEN
1) C MUST BE EQUAL TO MOD. OF A-B
2) C MUST BE LESS THAN MOD. OF A-B
3)C MUST BE GREATER THAN MOD. OF A-B
4)C MAY BE EQUAL TO MOD. OF A-B
Take your right hand fingers along A and curl these fingers along B, thumb will give you an idea for A × B. for the given problem answer will be west.
|A+B|2 = |A|2 + |B|2+ 2A.B (1)
|A-B|2 = |A|2 + |B|2 - 2A.B (2)
on subtracting (2) by (1)
|A+B|2 - |A-B|2 = 4A.B
so C2 = |A-B|2 + 4A.B
or C2 = |A-B|2 +4|A||B|cos120
since cos120 is negative so C will be less than |A-B|
Q A PARTICLE START MOVING FROM REST STATEVALONG A STRAIGHT LINE UNDER ACTION OF A CONST. F & TRAVEL DIST. X IN FIRST 5 sec THE DIST. TRAVELLED IN NEXT 5 sec WILL BE=?
QA CUBE OF A SIDE a IS HEATED TO RAISE ITS TEMP. BY 3*C IF Y IS COEFF. OF VOL. EXPANSION THEN TOTAL INCREASE IN AREA=? (ANS 12Yasqure. )
WAVES & OCILL. Q OBSV. IS MOVING TOWARDS A SOURCE AT REST WITH 30m/s if sound is sounding a sound of freq 500Hz & speed of sound is 300m/s then wavelenth by obsv. =?
Q TEMP. OF GAS INCREASED BY 2K KEEPING PRESS. CONST. SPEED OF SOUND IN GAS=? ans=1.22m/s
Q THE MASS OF ELEVATOR = 4000kg .STARTING FROM REST HOW MUCH DIST. WILL IT RISE IN 2s IF TENSION =56 kN . I want to know why we use ,' a ' for cal. dist. why 'a+g' is NOT use.
by formula s = ut + 1/2 a t2
u = 0 [given]
so s = 1/2 a (5)2= 25a/2 for first 5 sec.
and s = 1/2 a (10)2 = 50a for total 10 sec. from start
so for next 5 sec. distance = 50a - 25a/2 = 75a/2
answer : a'3 = a3 (1+ γ3) here a' is the side of the cube after heating it
so a' = a (1+3γ)1/3
so area increment = 6a'2 - 6a2 = 6a2 [(1+3γ)2/3 - 1] = 6a2 [ 2γ] = 12γa2
Please submit all the questions seperately.
λ = v/n = 300/500
i think you are asking about increment in speed
so according to formula v'/v = [(T+2)/T]1/2
or v'/v = 1 + (1/T)
so v' - v = v(1/T) = 332*1/273 = 1.22 here 332 is the speed of sound at 273K
when a machine works , you should not count gravity, for example if we raise our hand upword with uniform velocity then gravity makes no problem in it because our hand is working by our internal capabilitites
Q A PARTICLE STARTS MOVING WITH ACC.=2m/s 2 DISTANCE TRAVELLED BY IT IN 5th HALF SECOND IS=?
Q A PARTICLE MOVES IN A STRAIGHT LINE & ITS POSITION X AT TIME t IS GIVEN BY X RAISED TO P. 2=2+t . ACC. IS GIVEN BY=? (ans . -2/4t cube)
first calculate distance of 4 seconds by using s = ut + 1/2 a t2
then distance of 4.5 sec. by same formula
difference of these 2 is the distance of 5th half sec.
but some writter consider 5th half as 1/2 +1/2+1/2+1/2+1/2 means 2.5
so they use the same formula first for 2 sec then for 2.5 sec, then their difference.
according to my opinion first is more true way to solve.
Dear Sarika please submit only one question in a box, for a new question use another box so that i could give answers more quickly.
According to the given condition
x2 = 2+t
on differentiaing we get 2x dx/dt = 1
or v = 1/2x
now differentiate it again we get dv/dt = -(1/2x2 )(dx/dt) = -(1/2x2 )(-1/2x) = 1/4x3
A body is moving towards north with initial velocity of 13m/s. Its is subjected to a retardation of 2 m/s2 towards south.The distance travelled in 7th sec is ??
displacement of any particular sec. = u+at-(a/2)
here direction of motion is opposite to the direction of acceleration so formula will be u-at+(a/2)
displacement in 7th sec = 0
now we know that direction of motion will become south after 6.5 sec. so distance for last half sec = ut+(1/2)at2 = 1/4
same will be the distance for the first half sec of 7th. so total distance for 7th sec = 1/4 + 1/4 = 1/2