Messages posted by ~Áß???????

as U see in above diagram...when suppose boat is moving with vel = 8lm/hr in any direction @ from

the coast....only vsin@ term of V which is in x direction ..helps in crossing the river..

VCOS@ term is in y direction against water current

so for crosssing river in shortest time,

the vel Vx shud be max

so VSIN@ SHUD Be maximum

so @= 90degrees

HENCE PROVED!!

BEST OF LUCK

WELL IN THAT QUES TOO MCH INFO IS GIVEN ,, U SHUDNT HV GIVEN U..THERE...WE CUD HV DONE THIS WITHOUT U ONLY!!!

PE = kx and

then that kx = KE of ball...

18N/cm * 12 = 216 N

and nw KE = 1/2 * 0.015V^2

SO V = 170 APPROX I GUESS.....

BEST OF LUCK!!

when pilot was falling acceleration was in downward direction ....bt as soon as he opens that parachute net acceleration is in upward direction now...due to air resistive force.

IS C = R/ROOT(2) the answer?

WELL,indeed U CAN DERIVE THAT ALSO...

Sn = un + 1/2an^2

where n is time

Sn is distance after n secs..

and Sn+1 = u(n+1) + 1/2a(n+1)^2

so distance in nth second is Sn+1 - Sn = un + 1/2(2n+1)

@yash Maurya its nt (2n-1) i guess....

BEST OF LUCK

WHILE STOPPING ..OR APPLYING A BRAKE WE PUSH GROUND TO DECREASE OUR SPEED THERE...SO IT IS DUE TO REACTION FROM GROUND.

LET 2 CLIFFS HAVE DISTANCE BETWEEN THEM = L

AND BOY IS STANDING AT DISTANCE X FROM 1 CLIFF...

REFER FIG..

| |

|<------------------( X )---------------------> <----------------------------( L-x )------------------------>|

| (BOY) |

| |

( A) | | (cliff b)

lolz...(i m too lazy to make a attachment..)

so now fr travelling a distance 2X (AND HAVING A ECHO PHENOMENA) sound takes 1.5sec..

2X...bcoz to hear an echo sound have to come back from wall...lol

and for second echo from cliff B

2(L-X) = 2.5sec..

so X= 1.5 * 332 /2 = 249m

and L-X = 2.5 * 332 /2 = 415

SO TOTAL DISTANCE BETWEEN CLIFF = 415 m

MOREOVA 3RD ECHO HEARD AT TIME WHEN SOUND VOICE AFTER REFLECTION WILL AGAIN REACH BOY..FROM ANY 1 CLIFF...

SO IT WILL BE 2.5 + 1.5 = 4SECS...(TIME FR THIRD ECHO...)

BEST OF LUCK

IN THE FREE FALL...FORCE IS OBVIOUSLY ZERO AS BOTH ARE FALLING.....SO ...IN THAT FREE FALL STATE....relative accn of box = 0and yess...wen he lands and balances force increases greatly as change of momentum takes place....as m dv = F dt...so dv is very large ..towards end...and dt bcomes more smaller....so dv/dt=a bcomes large...and hence force increases....

well.....i m afraid u are wrong in that....according to newtons third law....THE WINNING TEAM AND ROPE pulls each other with equal force...when the winning team apllies force...there force is obviously greater than the losing team...and thats why because of f=ma....they pull the whole rope towards them including other team...LOLso the forces applied by winning and losing teams arent equal...(LET FORCE APPLIED BY WINNING T = FWAND BY LOSSING T = FLSO wen rope is in equilibrium in posn and none f em is moving...thn...FW=FLbt wen FW > FL...THE ROPE moves and resulting in displacement)There is an equal amount of force as FW which pulls wining tream also bt that is apllied by rope...instead f other team...which winning team withstand by ground friction....I HOPE U GT D POINT

well mass per unit length...or mass per length concept is veryy usefull in finding centre of mass of a body...It describes how the mass of a certain body is distributed according to its length..well in beginner level..u have mass per unit length as a numerical value like 2kg/m that signifies in evry 2m of that body .....mass is 2kg...here varying parameter is only length....assuming thickness and width of body as constant ..or nt considered...till IIT this application is enough...but for higher probs ...where we have to find moment of inertia and ....many torques and internal forces ....mass per unit legth can be in form of a quadratic eqn....or a binomial one also...like...9x^2 + 4x = 1....(just a e3xample...where x is distance from a defined origin point)better nt go into this much..JUST REMEMBER...mass per unit length describes distribution of mass....of a body as we go along its length...IT IS VERRYY IMPORTANT!!AND REMEMBA...finding centre of mass of a body IS inportant...so study it nicely...!!

is answer 1200 kms???moreova i have assumed that jouney have 2 parts ..n both are equal....

A) A STRAIGHT LINE.....as the person dropping coin is in d car...so both have same horizontal vel...so relative vel in horizontal motion is 0 ,,B) ofcourse stationary as explained above.

CONDITION FOR PURE ROTATION ..IS ..

AT THAT POINT OF TIME...FOR THAT OBJECT.

V= WR

V= LINEAR VEL

W= ANGULAR VEL

R = RADIUS...

AND FOR TOPPLING...

THE TORQUE ON BODY SHOLD BE ABLE TO ROTATE THE BODY...AT THE SAME TIME...

THE FRICTIONAL FORCWE SHOULD BE ABLE TO RESIST THE HORIZONTAL FORCE APPLIED...

MAKING IT STAY ..STILL ON THAT POINT OF TOPPLING

BEST OF LUCK!!

i think its ...root of (gH)

BEST OF LUCK!!

HERE...considering the figure

let ..2 bodies be point A and B

A is moving down ...and ..B is moving rightward direction.!!

now...considering figure..

initial distance = 40km..

after time T.....AA` = 20T....and.....BB`=40T

so ..S = ...root(2000T^2 - 800T + 1600).................(1)

NOW ..differentiating eqn 1 to zero.

u can get time T ..

HOPE U CAN SOLVE THAT EQN

I M TOO LAZY FOR THIS....

SORRY..

BEST OF LUCK

FIRST..taking ...on Incline Plane...

-->when it is case of pure rolling ....

then the frictional force is in upward direction ...in both cases (..i.e going down or up...)

-->when .....the body is sliding and going down or up(going up is rare)...frictional force direction is in opposite to..... the motion of the body!!!

-->and when body is slipping and going down or up(going up is rare) ....frictional force direction is in same direction ..that of motion of body!!!

SECOND ...taking...on flat surface..

-->in rolling and sliding condition ...frictional force ...in opposite direction of motion..

but in rolling case...the work done by friction is zero...and in sliding work done > 0

-->in slipping condition ..the frictional force in same direction of motion..!!

BEST OF LUCK

Initially mass = 0.5kg

height = 2m , g = 10m/sec^2..penetration = 5cm = 0.05m

so ....here let resistive force offrered by ..mass = 5kg...

IS EQUALTO R ..

so ...mg(h+x ) = Rx

0.5 * 10 * 2.05 = R*0.05

SO R = 205N....

now..velocity = 10m/sec..

so penetration...x = 1/2*0.5*100 / 205

x = 12.19 cm approx.....

now if box is free to penetrate....

then .. 0.5*10 = 5.5 V

V = 10/11m/sec.....

so here ....total ..energy conservation..

0.5*0.5*100 = 205x + 5.5*0.5*100/121

so x = 11cm...aprox..!!...

HOPE ANSWER IS CORRECT..!!

BEST OF LUCK..!!

here .....whatever the gas molecule be

the difference in Cp and Cv of gas molecule is always

R = 8.314 joules

so ....for 4 gm difference is 8.314J

FOR 1gm = 8.314 / 4

so = 2.8joule approx

now by conversion j = 4.27joule/cal

we can find heat in cal..!!

BEST OF LUCK

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