E - iR - L(di/dt) = 0

so nobdies got guts to answer

i dont really think u have answered correctly since the di/dt which is given includes the effect of the inductor and so does the final current which flows through it....

and even if u are correct and constant current flows due to the inductor what is the need to derieve a complex expression like

we could directly say the current is constant and would remain constant and is independent of the time....

given the following L-R circuit...

A battery of emf E is connected in series with a resistance R and inductor coil with inductance L...

now if i current flows through it then using kirchoff's law

E - iR - L(di/dt) = 0

and also gien di/dt = 5 A/s

question if we solve the eqn we'll always find the value of i to be same....however we are given di/dt=5A/s...

i.e the current is increasing at 5 A per second....

but the above equation gives constant value of i....

justify...........

ratings ensured........... ( mail me to know the answer)...

If u really wanna know ill tell you a book which the best ive ever read...

I ve read halliday, hc verma,dc pandey.....

but i found this one the best......concepts will be crystal clear.....

its a book on pfysics by JOHN TIPLER.....

yu can always check it out before buying

i dont thin it is possible beacause.... imagine the situation that the body is held by just one field above the ground at a place.... u wont see the ground below it rotating away and the body falling to the ground after sometime...

path of a projectile is a parabola...

u can find its eqn by following method....

let the particle be projected from ground with veleocity 'u' at an angle $\Theta$ with the horzontal...

its component in x dirn. = ucos $\theta$

in y dirn = usin $\theta$

now acceleration in x = 0...

in y = -g ( taking positive in upwards drin...)

let the particel be a at a point (x,y) at time t...

distance travelled in x = ucos $\theta$ .t ......................................(1)

distance travelled in y = usin $\theta$ .t + 1/2 (-g)t^2

= usin $\theta$ .t - 1/2gt^2 ....................................... (2)

from (1) t= x / ucos $\theta$

replace t by this in equation (2)

and ull get the eqn of parabola as

y = x.usin $\theta$ / ucos $\theta$ - 1/2g ${(x / ucos\theta)}^{2}$

.....

rate me if it helped and correct me if im wrong...

Ans 2...

il not use the concept of pseudo force as it ill confuse u....

no if the the ceiling move up with acceleration then the mass m will aslo move up with the same acceleration i,e

2ms^2

now accelration is upwards ..... kx is upwards.....and mg is downwards....

net force upwards give acceleration upwards.

kx - mg = ma....( again +ve dirn is upwards...)

put the values ( mg = 20N .... k = 100N/m...... m = 2kg.....a = 2ms^2....) an get the answer as....

100x - 20 = 4

100x = 24

x = 0.24 m....

::::::::::::::::::: Rate if u liked the answer ::::::::::::::::::::::::::

a)

let us assume that the spring is extended by elongation x.

when in eqillibrium state the net force on bock would be zero... refer to the fbd below....

forces on the body are :::

1. mg downwards....( 2 X 10 N in this case)...

2. Kx upwards ...... ( as spring exerts force in the opposite dirn of extension so as to .... restore its length..)

so... net force is 0 in equillibrium state....

mg - kx = 0 .....taking +ve in downwards dirn..

mg = kx

20 = 100 X x

x = 1/5 m

= 0.2 m

>>>>>>>>><<<<<<<<<

b) Forces in this case are ...

1. Mg downwards ....( note that M = ( 2 + 1)kg.... therfore Mg = 3 X 10 N)

2.kx' ( note x' is different from x )

applying the same concept as above and making net force 0

Mg = kx'

30 = 100 x'

x'= 0.3m

<<<<<<<<>>>>>>>

P.S ::::: Rate me if my answer helped....

The time period will increase beacuse when the child stands the effective length will decrease as the length will be measured from the point of suspension to childs centre of mass which wil be at a greater height when the child stands....

and we know $T = 2\pi\sqrt{L/G}$

Lets consider mass 2m hanging on the pulley...

now the pulley is massless...
therefore the net force on the pulley is 2T..... upwards and T' due to mass 2m via the string...

therefore, net force on pulley is 0... 2T = T'.......(1)

also from the figure we can easily see....

2mg - T' = 2ma'

2mg - 2T = 2ma'

also on the other block of mass m...

T - mgsin30 = ma...
T - mg/2 = ma

now we know that work done by internal tensions is 0...

if displacement of block of mass 2m in downwards dirn is x' and that of other block in direction of T on it is x

2Tx' - Tx = 0
2x' = x
double differentiating wrt t

2a' = a

put in the above equations and get the answer...

rate me...

the roof of the elevator is not as such visible in the figure...

however as we know the rope is at 2 pts of contact with the elevator and man system...

so at each of the pts one T acts...

let the two spheres have masses m and radii r and r' ( r----solid sphere,,.. r" --- hollow sphere)

for solid sphere
MI = 2/5 mr²

for hollow sphere

MI = 2/3mr"²

2/3mr"² = 2/5mr

r"²/3 = r²/5
r"/r=

(3/5)

pls rate if it helped...

well... ofcourse i acknowledge anchitsaini's effort... gr88

Let acceleration be a;;
mass of coin- m
now with elevator frame of reference
pseudo force = ma acts on coin in upwards dirn/...

and mg also acts on the coin in downward direction...

mg - ma = ma'

where, a' is the acceleration with elevator frame of reference

6 = 1/2a't^2
6 = 1/2a'
a' = 12 ft/s^2

put in earlier eqn

mg - ma = ma'

g - a = a' (dividing by m)

32 -a = 12
a = 20 ft/s^2

rate me if it helped...

i hope that u might have figured it out...

what i found was v and not v_{0

U HAD TO FIND V0 ON UR OWN USING EnRGY CONSERVATION....

(..look into the QS it asks for v initial and i am not so foolish so as to forgot what i have asked in the Q and give the answer myself...)

AND WHAT I MEANT BY I KNOW ANSWER WAS.. THAT ITS NOT MY DOUBT AND I JUST WANTED U PEOPLE TO TRY AND TEST.. URSELF..aND NOBODY ASKED U TO POST THE ANSWER LIKE THIS AND SPOIL THE FUN OF others wo are trying to to use their brains... That is why i gave a part of the answer and not the complete answer...}