Heres how u do the problem...
First let me tell u the basics of a transformer. A transformer is a mutual inductance based device that transfers power from one part to the other. It does so by virtue of stepping up the voltage or the current.
i.e VI = constant. So if the input voltage...
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Why don't u consider making ur own Gyroscope...and spin it while sitting in a chair such that its axis of spin is perpendicular to the ground. Now make it parallel to the ground . U will find that the chair now rotates
or you can look at the problem like this...
Total Energy is conserved in both the regions of the rope. Thus, consider the wave from the lighter density rope as the incident wave. When this incident wave meets the higher density rope, a part of it gets reflected and a part of it gets...
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What you should do is equate the electric potential energy to the kinetic energy of the electron...Thats all that you need to do and lo behold you have the answer.
i.e.
(e^2)/(4\Pi \epsilon_{0}r) = 1/2*(mv^2 )
From the above equation you can find the velocity...
Sridhar
Heres how u do it...
Since the mass of the bridge is concentrated at its centre, you can assume this to be a problem where the bridge is a single mass (do not consider length initially) connected to a rope making an angle of 30 degrees to the horizantal with a mass of 110 kg connected to the...
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Heres how u do the problem...
\int dw = \int_{x_{1}}^{x_{2}} F.dx; where F is the force and is equal to m*a, where m is the mass and a is the accln.
To find the work done by the force in 1st 4.4 seconds you will have to do this:
x = a + bt + dt^2 + et^3 . Therefore:
dx =...
No the accln is not and will not be greater than g:
This is because, the weight of both the bodies will act downwards while the accln on the lighte body due to the heavier body will act upwards. (Get the picture?) Hence the force eqtn wil be like this:
m_{1} * g - T = m_{2}*(g-a)
So...
I'll just give you a small hint. The Masses are suspended from the pulley on either side of the uplley. So the heavier mass is going to exert a force F on the lighter mass and will try to make it move towards itself. This force exerted by the heavier mass - tension will be equal to the force on...
You can also do something like this using Latex:
\int_{0}^\infty f(x)\partial x
or something like:
\sum (n) = n(n+1)/2
So as you see the display beomes more organised wit latex. Yet the time taken to type an answer using latex is enormous..
Sridhar
tex is basically the latex tag that is used to begin typing latex commands that can be used to generate symbols like I have in my 2nd answer. I screwed up on the tags in the previous answer, so I have corrected them in my second Just click on one of those tags to see how the symbols are...
Just a repeat answer since the symbols didn't come out correctly ...
The Work Done = Change in Kinetic Energy, which means that:
F * d = m/2(v_{f}^2 - v_{i}^2),where,
v_{i} & v_{f} are the initial and final velocities respectively and F is the force and d is the displacement. In your case...
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Assuming that the mass of the galaxy is concentrated at its center, the force of the galaxy of mass M on the sun of mass m separated by a distance r from the galaxy center is:
F = GMm/r^2. This force will be equal to the centrifugal force due the revolution of the sun around the galaxy...
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Let F[/f]be the force exerted by the falling ball of mass m on the spring with spring constant k. Since the ball is under free fall, the force wih which the ball falls on the spring is its weight mg. Hence,
mg=-kx, where xis the displacement of the spring.
Therefore, [b]x=mg/k. After...
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You have answered ur question... "c" is the ultimate speed. Though its value is finite it has got infinity like properties: c+c=c, c+1 = c etc.Special Relativity also says that C is constant with respect to all reference frames. Which means whether u are moving, stationary, accelerating, C...
Hi
Heres how u go about with the problem:
the vertical component of the acceleration is along the normal force and the horizontal component of the acceleration is along mgsin28 component. Thus a equilibrium, if a = min accln req by the mass to slide down:
μ*(masin28 + mgcos 28) =...
Hi
Ok, the sling shot technique is one where u use a fully stretched rubber band. However I have been calculating a few things. You would get a better range if u used a spring technique. Here are the procedures:
1) Fix one end of the spring to a vertical rod that stands on the ground...
Hi
Have u tried the Sling Shot technique? or have u tried filling the can completely with pressurised soda/gas and then opening the cork so that the can moves forward? or u could even use a spring
Sridhar
Hi
According to what I understood of the problem this is how u proceed:
Let T = tension, m = mass of the cork, v = linear velocity of the cork, r = radius, f = frequency, w = ang.frequency; a=accln
then,
T - (mv2)/r = M*a
w = 2*π*f
Substitute this and find the value of T w.r.t f...
Reformatted reply,
Since the answer was not formatted properly, here's the same answer formatted again...
This is how u go about with the problem:
First calculate the vertical velocity of the stone using
v = √(2*g*h) and then calculate the time taken t for the ball to reach the ground...
Hi,
This is how u go about with the problem:
First calculate the vertical velocity of the stone using
v = & radic ;(2*g*h) and then calculate the time taken t for the ball to reach the ground vertically, then using the horizontal velocity that is given to be = 14m/s by using t = 75.7/vand...
My previous reply, more organized...
Hi, since my previous reply did not contain thetas...i am re formatting my reply here...
The situation is similar to a box lying on an iclined plane and finding the angle of inclination at which the box will start sliding. As the ramp is lifted, the box...
Hi
I don't agree with Hallsof Ivy. The situation is similar to a box lying on an iclined plane and finding the angle of inclination at which the box will start sliding. As the ramp is lifted, the box stays in position as long as &mu*mg*cos&theta = mg*sin&theta.(&theta being the angle of...
Hi,
You have not considered the Centre of Mass of the system.
Consider the mass of the man to be m, mass of the boat to be M,
velocity of the man to be u and velocity of the centre of mass of the
man-boat system to be v.
The velocity of the boat gets adjusted to the velocity of the man...
Try this...
In any elastic collision, the linear Momentum and Kinetic Energy are conserved. Thus if v = final velocity of 3m, then
v = [(2*m)/m+3m]*9.81 = 4.905 m/s
Hope u have ur problem solved...
Sridhar
The solution
Hi,
This is the soln from what I understood of the problem...
Just equate kx = mrw2
w = 25 rev/s
m = .15 kg
r = 0.06m
x = 0.04m
U will get the value of k now...
How this helps u...
Sridhar