Dear swetlina,

When a rocket is fired then as it moves verticaly upwards its kinetic potential energy increases at the expense of its kinetic energy. and it acquires maximum height at a point where its potentil energy is equal to its initial kinetic energy.

However, due atmospheric resistance offered, only 80 percent of its kinetic energy is getting converted into potential energy

so use the formula

(80/100)(1/2m*2000*2000)= maximum potential energy

please take a trouble of writing problem instead of giving references, as the sequence of problems in a book differs depending on the its edition.

However ur problem is illustrated as follows:

mass of boat = 50Kg

volume of the boat = 1 cubic meter

To find fraction of the volume of the boat immersed in water

we make use of Buoyancy principle

buoyant force = wt. of the boat

(mass of volume of the water displaced) * g= 50*g

1000*V = 50

here density of water is 1000kg/m^3

therefore V = 1/20

so fraction of volume of boat immersed in water = 1/20

Time taken by the ball to reach ground = 2 sec

when the ball was projected vertically upwards then its horizontal velocity = horizontal velocity of car at the moment when the ball was released = u (say)

so, horizontal distance travelled by the ball in 2 sec = ut = 2u meters ...... (1)

(Remember here there is no component of acceleration for the ball in horizontal direction once it is released)

Distance travelled by the car is given by

s = ut + (1/2)at²

    = 2u + (1/2)*1*(2)²

or s  =2u + 2    ............(2)

From (1) and (2)

The distance at which ball will fall behind the boy on the car is

= (2u + 2 ) - 2u = 2 meters

There are certain intersting paradoxes and consequence in special theory of relativity as proposed by Einstien. These are

1) Ladder paradox

2) Relativity of simultaniety

3) twin paradox

4) length contraction

5) Time dilation

The twin paradox, stems from special relativity: one of two twin brothers undertakes a long space journey with a high-speed rocket at almost the speed of light, while the other twin remains on Earth. When the traveler returns to Earth, he is younger than the twin who stayed on earth.

This appears to be a paradox if one expects that, according to relativity, either twin may validly claim to be "at rest", and thus each expects that the other twin will age slowly. In this case the twin who turns round pursues two different paths through space-time whereas the other twin has a constant linear path. Either twin could verify this by jettisoning material at intervals (the material would be arranged in a line for one twin and form two lines for the other) or by observing onboard accelerometers.

Consider a space ship going from Earth to the nearest star system a distance d = 4.45 light years away, at speed v = 0.866c (i.e., 86.6% of the speed of light). The round trip will take t = 2d / v = 10.28 years in Earth time (i.e. everybody on earth will be 10.28 years older when the ship returns). Ignoring the effects of the earth's rotation on its axis and around the sun (at speeds negligible compared to the speed of light), those on Earth predict the aging of the travellers during their trip as reduced by the factor , the inverse of the Lorentz factor. In this case ? = 0.5 and they expect the travellers to be 0.5×10.28 = 5.14 years older when they return.

The ship's crew members calculate how long the trip will take them. They know that the distant star system and the earth are moving relative to the ship at speed v during the trip, and in their rest frame the distance between the earth and the star system is ?d = 0.5d = 2.23 light years ("length contraction"), for both the outward and return journeys. Each half of the journey takes 2.23 / v = 2.57 years, and the round trip takes 2×2.57 = 5.14 years. The crew arrives home having aged 5.14 years, just as those on Earth expected.

If a pair of twins were born on the day the ship left, and one went on the journey while the other stayed on earth, the twins will meet again when the traveller is 5.14 years old and the stay-at-home twin is 10.28 years old. This outcome is predicted by Einstein's special theory of relativity. It is a consequence of the experimentally verified phenomenon of time dilation, in which a moving clock is found to experience a reduced amount of proper time as determined by clocks synchronized with a stationary clock. Examples of the experimental evidence can be found at Experimental Confirmation of Time dilation.

The strange result of twins of different physical age was not the central problem of the "twin paradox". As early as 1905, Einstein predicted that a clock which is moved away and brought back will lag behind stationary clocks. Einstein called that result "peculiar", but the calculation is straightforward and the example was not presented as paradoxical

The usual resolution of the paradox as presented in physics text books ignores its origin (it only surfaced with general relativity, see above) and regards it as a problem due to misunderstanding of special relativity. Here the Earth and the ship are not in a symmetrical relationship: the ship has a "turnaround" in which it feels inertial forces, while the Earth has no such turnaround. Since there is no symmetry, it is not paradoxical if one twin is younger than the other. Nevertheless it is still useful to show that special relativity is self-consistent, and how the calculation is done from the standpoint of the traveling twin.

Special relativity does not claim that all observers are equivalent, only that all observers in inertial reference frames are equivalent. But the space ship jumps frames (accelerates) when it does a U-turn. The twin on Earth rests in the same inertial frame for the whole duration of the flight (no accelerating or decelerating forces apply to him or her) and he is therefore able to distinguish himself as "privileged" compared with the space ship twin. The accepted resolution of the paradox is that a calculation different from that above must be made for the crew, a calculation which explicitly recognizes the change of reference frame, and the change in simultaneity which occurs at the turnaround.

There are indeed not two but three relevant inertial frames: the one in which the stay-at-home twin remains at rest, the one in which the traveling twin is at rest on his outward trip, and the one in which he is at rest on his way home. It is during the acceleration at the U-turn that the traveling twin switches frames. That's when he must adjust the calculated age of the twin at rest. Here's why.

In special relativity there is no concept of absolute present. A present is defined as a set of events that are simultaneous from the point of view of a given observer. The notion of simultaneity depends on the frame of reference (see relativity of simultaneity), so switching between frames requires an adjustment in the definition of the present. If one imagines a present as a (three-dimensional) simultaneity plane in Minkowski space, then switching frames results in changing the inclination of the plane.

In the spacetime diagram on the right, the first twin's lifeline coincides with the vertical axis (his position is constant in space, moving only in time). On the first leg of the trip, the second twin moves to the right (black sloped line); and on the second leg, back to the left. Blue lines show the planes of simultaneity for the traveling twin during the first leg of the journey; red lines, during the second leg. Just before turnover, the traveling twin calculates the age of the resting twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnover, if he recalculates, he'll measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the lifeline of the resting twin. The resting twin has suddenly "aged" very fast, in the reckoning of the traveling twin.

 

Electomagnetic waves have got sinusoidal electric and magnetic field mutually perpendicular to each other and the propagation vector (poynting vector) is perpendicular to the palne containing electric and magnetic field. 

Electromagnetic wave does not require any medium to propagate unlike sound or material waves. It is because em waves have got changing magnetic and electric field. This chage is temporal as well as spatial in nature. Infact spatial and temporal change of magnetic field gives rise to electric field (faraday's law of electromagnetic induction). Further changing magnetic field gives rise to electric field. Thus, electric and magnetic fields are sustaining each other (though there is no electric charge or magnetic poles in the vacuum).

Thus, to summarise, the spatial and temporal change in electric field sustains magnetic field and vice versa, and by virtue of this property em waves does not require any medium to propagate. It can be explained in a very elegant manner using Maxwell' equations.

x²(x dx + y dy) + y (x dy - y dx) = 0

or (x dx + y dy) + y (x dy - y dx)/x² = 0

or d (xy) + d (y/x) = 0

or d ( xy + y/x) = 0

or ( xy + y/x) = constant (say C)

or ( xy + y/x) = C

or y = C/(x + 1/x)

or y = Cx/(x+1) which is the required solution.

here train is moving in upward direction on an inclined plane having angle of inclination = 30 deg.

acceleration of the train = a = g/2

here component of acceleration in upward direction i.e. normal to the horizontal is (g/2)sin30 and along the horizontal it is (g/2)cos30

let string makes an angle  with the normal to the celing of train.

Tension in the string = T

Resolving the tension along two components as follows:

(a) Vertical to the ground

T cos( - 30) = m[g +(g sin30)/2]     (1)

(b) Along the ground that is in the horizontal direction

T sin( - 30) = mg (cos30)/2     (2)

Dividing eq. (2) by (1) we obtain,

tan( - 30) = [mg (cos30)]/m[g +(g sin30)/2] = 3 /5

or (tan - tan30)/(1+tantan30) = 3 /5

or tan = 2/3

or  = tan^-12/3 

Now to obtain tension in the string square and add then equate LHS and RHS of eq. (1) and (2)

by doing so we obtain and simplifing we obtain

T = 57

Q) sec+tan=100,find the value of tan+cot

Since, sec+tan=100

or 1/cos + sin/cos = 100

or 1+ sin = 100cos

Squaring both the sides

(1+ sin )(1+ sin )= 10000 (cos)((cos

or (1+ sin )(1+ sin )= 10000(1 - sin²)

or (1+ sin )(1+ sin )= 10000(1 - sin)(1 + sin)

therefore cos =[(1 - (9999/10001)²]^1/2= (200/10001)              (2)

tan+cot = (10001)²/[9999×200]

Dear Rahul i didn't exactly understand the meaning of 'lowest polynomial'. However, i consider it as a polynomial of lowest degree and solve the problem as below:

(x^{2_+x+1)(x2_{-x+1) =}} x⁴ + x²+ 1

Thus, if we add x² to above term it becomes a perfect square i.e.

(x⁴ + x²+ 1) + x² = x⁴ + 2x²+ 1 = (x²+ 1)²

0/0 is not defined.

Infact, any real number divided by 0 is indeterminate or not defined.

Eg. 2/0, 5/0, 6.5/0 etc. can not be defined.

To put it more lucidly consider

8/2=4 its because 2 multiplied by 4 is 8

20/4=5 its because 4 multiplied by 5 is 20

in general,  x/y = z means 'y' multiplied by 'z' gives 'x'

Now, if you write 6/0 then can you imagine a number which on multiplication with 0 gives 6 !!!!!!!!!!!!!!!!!!

Well, it can't be 2, it can't be 5, 4, 89, 1989, 100000000 or any other number thus, anything divided by 0 is not defined (indeterminate)

 

Further if we come to 0/0, then it is again 'not defined'.

But you may argue that 0/0=0 because 0 multiplied by 0 is 0, then on similar grounds i may argue that it can be 1, 2,3,5 or any other number, as anything multiplied by 0 is 0. Thus nothing definite is coming out. So such argument is absurd and is just a way of explaining or understanding a concept.

Finally, to conclude with we say 0/0 is not defined.

When we say that in concave mirror 1/u + 1/v = 1/f

then here u, v, and f are absolute values and further derivation is done by assuming a particular ray diagram in which object is assumed to be at a particular position (though fortunately the final expression comes out to be same irrespective of various positions of an object) So, it is necessary to take care of sign convention while solving a numerical too. You can further have an insight for the explanation by solving the mirror or lens problem not by directly putting values of u, v and f in the formula taking care of sign convention, but by deriving the expression again using given numerical values taking care of sign convention. I hope that i am able to clarify your doubt.

(6 6 + 14)²ⁿ⁺¹= p

Here, to get the integral part we need to consider only the additon of those terms in which the power of 6 6 is even, as even power of 6 6 will be an even number and this when multiplied by some power of 14 will again give even integer and finally addition of such even terms will also be even.

As far as second part is concerned, it seems that there is problem or error in it for the reason that you have mentioned that p(f), where f is a fractional part and you are asking to prove a fraction part equal to 20²ⁿ⁺¹ which in turn comes out to be an even integer. So there is a discrepancy in the problem. Kindly check the same

As Mr. kaushalsapre  has very rightly remarked that car moving with constant velocity is an Inertial frame of reference and hence, according to the law of Relativity, " in an inertial frame of reference laws of physics are exactly the same", i.e. all experiments conducted in inertial frames will give exactly identical results and thus it is not possible to find whether the car is moving or not. However, an accelerating body can be taken as non inertial frame of reference in which the results of experiments conducted will be different. for example in a car accerlerating with some acceleration 'a' the string to which bob is attached will not point in the vertical direction but it will subtend some angle with the normal drawn from the point of suspension. Earth is also an inertial frame of reference though it is rotating but rotational velocity is so small that it can be ignored.

thus, frames at rest or moving with constant linear velocity are inertial frames of references. Whereas rotating and accelerating frames are non inertial frames of references.

_{SINCE,  0 ]}^{[1 ]} f(x) dx = 1,  _{[0 ]}^{[ 1]} xf(dx) = a  &  _{[0 ]}^{[ 1]} x² f(x) dx  a², 

therefore, I = _{[0 ]}^{[ 1]} (a -x)²f(x) dx is = _{[0 ]}^{[ 1]} (a² - 2ax + x²)f(x) dx

           or I = _{[0 ]}^{[ 1]} a² f(x) dx -2a_{[0 ]}^{[ 1]} x f(x) dx + _{[0 ]}^{[ 1]} x²f(x) dx

or I = a² _{[0 ]}^{[ 1]} f(x) dx - 2a(a) + a^{2 =} a².1 - 2a.a + a²

so, I = 0