Messages posted by annu93@rediffmail.com

centrifugal force is a pseudo force which we take into account when we discuss dynamics of particle moving in the circle from that particle frame.

numerically = mv²/R

tim of light of the con is the time of motion od the car

time of flight = 20/ 40 hr = 0.5 hr .

distance covered by the con = 0.5*100 = 50 Km

Gaan's graph is perfectly correct.

when we have assumed downward as positive then sign convention does not change during the whole motion .

i would say if you are searching for a set of a long list of very hard questions that you willl not find in VMC.

to be more precise mass of 1 mole of alpha particles is 4g .

AM = a+b/2= 8

=> sum of roots = 16

G.M = (ab)^1/2 = 5 => ab = 25 .

required quadratic equation

x² - 16 x +25

time = 0.69 s

the question has asked distance moved relative to the lift which is equal to = 2.7 m

yaar give us some dimensions as toh eight of loop player and some other .

if 5 unpaired electrons provide more stability then will a d⁶ lose a electron to achieve a half fillef configuration ?

2 log_x+1x²+x-6 = 4

log_x+1x²+x-6 = 2

(x+1)² = x²+x-6

x² + 2x+1 = x²+x-6

3x = -5

x = -5/3

also base of the logarithms x+1>0

=> x>-1

thus i think x has no solution

force is defined as "rate of change of momentum "

in calculas

f = dp/dt

= d*m*v/dt

= mdv/dt

= ma

when all the terms are used in S.I units then value of constant k becomes 1

it's very simple .

weight = mass * g

=> 9.8 = m * 9.8

=> mass = 1 kg

now when force = 20 N

a = F/m = 20/1 = 20 m/s²

gor one you can't determine the constant (if any) .

and a dimensionally correct formula may not be a correct one.

I am getting

vy = ux ( taking A as the origin )

you haven't specified the direction of the flow of the river so i hae taken along the banks of the riiver. i.e perpendicular to AB

.

from the diagram we can see that the angle between normal and gravitational force decreases .

thus component of g-force normal to surface increases unitl at the vertex it becomes same.

both cars cover angular dispalcement of 360 degrees .

and it is given that t₁ = t₂

thus angular velocity for both is equal.

acceleration and force on each car differs.

both cars cover angular dispalcement of 360 degrees .

and it is given that t₁ = t₂

thus angular velocity for both is equal.

acceleration and force on each car differs.

relative velocity of aprticle A w.r.t to particle B

time taken to coliide = 100/ 40

= 2.5 s

b) for paticle dropped from 100 m

s= gt²/2

s = 10*6.25/2

s = 31.25 m

point of collision above the ground = 100 - 31.25 = 68.75 m

Z² is not necessary fopr hydrogen but all other elements it has to used .

let Z = a +ib X = a-ib

|x|² = ZX

=> K = 1

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