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 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Wt of the cylinder=Wt of the liquid displaced= 2(pie)rtps (r+l)g

Volume of the liquid displaced =Volume of cylinder submerged= 2(pie)rtps (r+l)/ pl

Total volume of cylinder= (pie) r2

Hence

f=Volume of cylinder submerged / total volume of cylinder

f= [2(pie)rtps (r+l)/ pl ]/ (pie) r2 l

f=2tps (r+l)/pl rl

 Discussion Forums -> This Post 7 points    (1    in 2 votes )   [?]

Applying conservation of momentum:

Hence velocity of combined body V= u/2

where u is velocity of sphere before collision:

Total energy required to be imparted to the system to make full rotation:

2mgl=1/2 (2m)(u/2)2

u=2 root(2gl)

 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]

All have solved it correctly:

Now let us explain it in my own way

The man feel his wt. because of normal reaction on his feet. Let us consider downward direction +tive

There only two forces acting on the man

1. Normal reaction "N" upward

2. Wt i.e. mg downward

Hence  Summation of all forces= ma

mg-N=mg

N=m(g-a)

Hence result

Now in second case:

only direction of accn has changed:

Hence a in above equation must be replaced by  (-a )

N=m(a+g)   = W/g (a+g)=W(a/g +1)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Wt of the cylinder=Wt of the liquid displaced= 2(pie)rtps (r+l)g

Volume of the liquid= 2(pie)rtps (r+l)/ pl

Total volume of cylinder= (pie) r2 l

Hence

f= [2(pie)rtps (r+l)/ pl ]/ (pie) r2 l

f=2tps (r+l)/pl rl

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

From above fig:

Let m=6kg, M=15kg

=300

N1=m(g-a1)

But a1=a2

Hence N1=m(g-a2 ) ..............(1)

For M:

Ma2=(N1+Mg)..........................(2)

From (1), (2)

a2=g(m+M)/{M+m2 }

a2=70/11 m/sec2

a1=35/11 m/sec2

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Hi

I am solving Erodov prob u have asked me

Let at time t the position of the particle is P(x1, y1) and at t+dt is Q(x1+dx, y1+dy)

ds=dx i + dy j where ( x1=R, y1=R )

ds/dt= -Rd i + Rd j

v= -Rd/dt i + Rd /dt j

Vav where limit for Theta is "0" to 1800

=-2R/T i + 0. j

|Vav | = 2R/T   This is average of velocity vector.

Mean velocity is easy to find:

Distance= Pie R Time =T

V=(Pie R) / T

,

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Initial height h is gained with uniform velocity v

Hence t1 = h/v .....................(1)

Now engine fails , hence its speed will be zero after further going h1 height in next t2 sec.

h1 = v2 /2g..........................(2)

t2 = v/g....................................(3)

Total height H= h+h1 =1/2 g t3 2  where t3 is time to descend

v2 /2g +h =1/2 g(t -t1 -t2)2  ...........(4)

Now substitute the value of t1 and t2 in equation (4) and solve for h

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Conside the figure sent by Sivam:

The triangle shows the corners that will contain Centre of mass of base spheres. Normal reaction(R)due to top sphere will make an angle = 1/

Tension T will make 300 with each of the bisectors of the base c.g. angles of triangle.

Hence  3R=2   (weight of top sphere)

3R( ) =2

R=2

2T=R=2/

T=2/3

Since safety of faxtor is 3

Tmax =2

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

Let at time t velocity of trolley is vt

mass of trolly=(m+ut)

momentum of the system=v(ut) + (m+ut)vt

At time t+dt, corresponding velocity of trolley vt + dvt , mass=m+u(t+dt)

momentum of the system=vu(t+dt) +(m+ut+udt)(vt + dvt )

Since there is no external forces acting on the system, hence momentum is conserved:

v(ut) + (m+ut)vt =vu(t+dt) +(m+ut+udt)(vt + dvt )

Dividing both sides by dt we get:

(m+ut)dvt /dt= -u(v+vt )

Solving above dirrerential equation for vt , we get:

(m+ut)(v+vt ) = K( a constant)

when t=o, vt =0

Hence k=mv

Hence vt = - uvt/(m+ut)

dvt / dt= accn = -muv/(m+ut)2

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Tmax=2

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

e=0.8415

M=16.475kg

I will explain if this answer is correct

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

When its accn :

(50)2 =100+2as

2as=2400............................(1)  where a=accn,  s=distance

When its direction of  acce is reversed, its velocity will first become zero then goes in reverse direction with accn  a

0=(50)2 - 2as1

2as1=2500...............................(2)

Hence velocity at starting point: Total distance =s+s1

v2=0 + 2a(s+s1 )=4900

v=70m/s Ans

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Let vt is velocity of trolley after t sec.

Hence (v- vt) u=mdvt /dt

Solving above equation you get:

vt=v(1- e-u/m t )   Ans.

Assumptions:

1. No dust accumulate in the trolley.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Hi Zeosthegod

Although you have cosen right options,

but you have solved incorrectly the question (ii)

because u/2i + u j= root of(5) X u/2

So use here energy equation to find velocity of the ball, then solve it

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

The resultant accn of block B is vector sum of  b and c. So add them vectorically and get the answer.

 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]

3/16 Ma2

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

I am trying to solve this:

Let distance travelled on the plane = d

Elongation in spring= dx=Square root of(l0 2 + d2)  - l0

Tension in spring= T = k(dx) = (5mg/l0 )(dx)

Now let spring makes an angle  with horizontal plane

Remember just at the time of taking off the mass from the floor (mg) equals the vertical component of T

Hence T =mg

(5mg/l0 )(dx){l0/root(l02 + d2)}=mg

But dx=Square root of(l0 2 + d2)  - l0   Solving for (d/l0)

We get : d= (3/4)l0

Now writting energy equation for the system we get:

k(dx)2 + (2mv2)  = 2mgd

Now put the value of d, we get

v= square root of (19/32 gl0 ) = 1.7m/s

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

nobody can understand this problem unless u attach the diagram or indicate the exact location of spring , masses and points  O, A and P

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

Hey

B, C and A

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Hi ABHI

In an elastic or plastic collision reduction in energy appears in the form of heat and some part is consumed in deforming the bodies in question.

So tell me where this energy is going? Is there any deformation ?

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