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I'm studying for the exam day after tommorow and while reading the practical on jaeger method I found the following... P1-P2=2T/ r r is the radius of the capillary tube .............. advantages of this method.., •contact angle is not needed •can find surface tension of liquids like...

Yes...so whats the physics behind that ? water drains from symetrically placed holes from the center will catch the ground at the same place....? all arcs are paths of water pouring.red color shows some special feature.. further...siphon needs a continuous flow of water.water bubbles disturb...

1. I found this diagram on book but there weren't any description.can someone tell me, what its trying to tell specially by those two red lines meeting the ground at the same place...? 2.this is a diagram for siphon method of removing water. I have read somewhere that the siphon stops if the...

Thanks a lot Cwatters !

I am trying to figure out what happens here...and now I've stuck on this... Here it goes...., situation 1 Both contact A with B and A with ground are rough We increase the force P applied here from zero.When there's sufficiently big μs between A and B than that of A and ground,there's an...

Tension in the rope is like this..(write click,view image to zoom) http://upload.wikimedia.org/wikipedia/commons/8/8f/Tension_figure.svg

hey..I'm not a British.. :)...Not even near it I've not confused this weight and mass.... You may have confused this... I have use the brackets to show that 1000N has came from 100kg mass. and also I've taken the gravitational acceleration as 10ms-2. The whole system can hold a mass of...

In both situation the tension of the rope is 1000N(100kg) Even in the second situation you will be able to hold only 100oN tension.You won't be able to double the maximum tension it can hold by bending the rope.But you can hold more masses by dividing the force created by the mass on rope,just...

Is it something like this ? -we just take velocity changing discontinuously mathematically though it changes continuously.Do we just dismiss what happens actually inside it? So how can we say that the velocity changes discontinuously ?

I think that the impulsive force generated by one ball on the other slows down to o ms-1 till it changes all its kinetic energy into elastic potential energy and speeds up to 2 ms-1 in the other direction changing all elastic potential energy into kinetic energy in milliseconds.I also thinks...

I was reading an article about velocity and displacement of a particle..It says "though the position of a particle is a continuous function of time,velocity sometimes isn't.For instance velocity is not continuous while an impulsive force is being applied." But I think that velocity changes...

You know the surface of the hard board is not regular(even)..

How can that be ?...we can have some mean value for that...when we take the hard board..there are +error and -errors...so the mean value will reduce the quantity of error

edited.... Types of errors 1.Systematic errors Zero Error errors due to slow stopwatches errors due to incorrectly graduated scales parallax error 2.Random errors error due to unevenness of the measuring item Ways of comparing errors Absolute error\rightarrowerror when...

let's take that we're going to measure the thickness of a hard board..well if the surface is uneven then we get a wrong value..but we can minimize the error by getting measure from more than one place and calculating the mean.. Why?..Isn't it a type of error..can't we divide errors into these...

Here's what I understand.... Types of errors 1.Systematic errors Zero Error errors due to slow stopwatches errors due to incorrectly graduated scales 2.Random errors error due to unevenness of the measuring item parallax error Ways of comparing errors Absolute...

Oh...sorry..I have written it wrong... f(x)=(x-α)g(x)+f(α)(remainder theorem) g(x)=(x-β)∅(x)+f(β)(remainder theorem) f(x)=(x-α)[(x-β)∅(x)+β]+f(σ) f(x)=(x-α)(x-β)∅(x)+f(β)(x-α)+f(σ) Here we get something like A(x-α)+B as the remainder{f(β)(x-α)+f(σ)} A=f(β) and B=f(σ) Correct now ...

Homework Statement Applying remainder theorem again and again to show that the remainder of the f(x) polynomial function when divided by (x-α)(x-β) is A(x-α)+B . Determine A and B Homework Equations the remainder of a polynomial f(x), divided by a linear divisor x-a, is equal to f(a) The...

thanks Infinitum and Pranav-Arora...Successfully solved the 2nd and 3rd limits..I'm going to ask about 1st one from a teacher...thanks for the help !

what about the [3] one?

We have not such rule in Limits but in differential calculus we were taught, y=f(x)g(x) \frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)

urgh..then its the derivative of ysec(y)....so then ? Is it the answer ?-ysec(y) ?

\frac{1}{\sqrt{2}} as cos π/4 or sin π/4 ?

OMG..sorry...it's not x to a ,but x to 0....sorry

congratz ! EDIT: Woot! 37 posts.

That's what I'm gong to do next.. Anyway what about the other two limits ? [2] and [3] ?

No..only these two are available in our coursework.. Nothing like following in ours \lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}

(1+0)1/0=1\infty=\infty ?

(1+0)1/0=1\infty=\infty ?

spends a lot to type Latex...It cost my time and yours