AS pendulum clock gives correct tiem at the equator, Wil it gain tme r loose time as it is taken to the poles?

A pendulum clock loses 12s a day if the temperature is 40^@C and gains 4s a day if the temperature is 20^@C , The temperature at which the clock will show correct time, and the co-efficient of linear expansion (alpha) of the metal of the pendulum shaft are respectively:

A simple pendulum is taken from the equator to the pole. Its period………..

A second's pendulum clock has a steel wire. The clock is calibrated at 20^@ C . How much time does the clock lose or gain in one week when the temperature is increased to 30^@ C ? alpha_(steel) = 1.2 xx 10^-5(^@C)^-1 .

Path traced by a moving particle in space is called trajectory of the particle. Shape of trajectiry is decided by the forces acting on the particle. When a coordinate system is associated with a particle motion, the curve equation in which the particle moves [y=f(x)] is called equation of trajectory. It is just giving us the relation among x and y coordinates of the particle i.e. the locus of particle. To find equation of trajectory of a particle, find first x and y coordinates of the particle as a function of time eliminate the time factor. The position vector of car w.r.t. its starting point is given as vec(r)=at hat(i)- bt^(2) hat(j) where a and b are positive constants. The locus of a particle is:-

Path traced by a moving particle in space is called trajectory of the particle. Shape of trajectiry is decided by the forces acting on the particle. When a coordinate system is associated with a particle motion, the curve equation in which the particle moves [y=f(x)] is called equation of trajectory. It is just giving us the relation among x and y coordinates of the particle i.e. the locus of particle. To find equation of trajectory of a particle, find first x and y coordinates of the particle as a function of time eliminate the time factor. In above question initial acceleration (i.e. (d^(2)vec(r))/(dt^(2))) of particle is :-

Path traced by a moving particle in space is called trajectory of the particle. Shape of trajectiry is decided by the forces acting on the particle. When a coordinate system is associated with a particle motion, the curve equation in which the particle moves [y=f(x)] is called equation of trajectory. It is just giving us the relation among x and y coordinates of the particle i.e. the locus of particle. To find equation of trajectory of a particle, find first x and y coordinates of the particle as a function of time eliminate the time factor. In above the velocity (i.e. (dvec(r))/(dt)) at t=0 is :-

Einstein in 1905 proppunded the special theory of relativity and in 1915 proposed the general theory of relativity. The special theory deals with inertial frames of reference. The general theory of relativity deals with problems in which one frame of reference. He assumed that fixed frame is accelerated w.r.t. another frame of reference of reference cannot be located. Postulated of special theory of realtivity ● The laws of physics have the same form in all inertial systems. ● The velocity light in empty space is a unicersal constant the same for all observers. ● Einstein proved the following facts based on his theory of special relativity. Let v be the velocity of the speceship w.r.t a given frame of reference. The obserations are made by an observer in that reference frame. ● All clocks on the spaceship wil go slow by a factor sqrt(1-v^(2)//c^(2)) ● All objects on the spaceship will have contracted in length by a factor sqrt(1-v^(2)//c^(2)) ● The mass of the spaceship increases by a factor sqrt(1-v^(2)//c^(2)) ● Mass and energy are interconvertable E = mc^(2) The speed of a meterial object can never exceed the velocity of light. ● If two objects A and B are moving with velocity u and v w.r.t each other along the x -axis, the relative velocity of A w.r.t. B = (u-v)/(1-uv//v^(2)) One cosmic ray particle appraches the earth along its axis with a velocity of 0.9c towards the north the and another one with a velocity of 0.5c towards the south pole. The relative speed of approcach of one particle w.r.t. another is-

Einstein in 1905 proppunded the special theory of relativity and in 1915 proposed the general theory of relativity. The special theory deals with inertial frames of reference. The general theory of relativity deals with problems in which one frame of reference. He assumed that fixed frame is accelerated w.r.t. another frame of reference of reference cannot be located. Postulated of special theory of realtivity ● The laws of physics have the same form in all inertial systems. ● The velocity light in empty space is a unicersal constant the same for all observers. ● Einstein proved the following facts based on his theory of special relativity. Let v be the velocity of the speceship w.r.t a given frame of reference. The obserations are made by an observer in that reference frame. ● All clocks on the spaceship wil go slow by a factor sqrt(1-v^(2)//c^(2)) ● All objects on the spaceship will have contracted in length by a factor sqrt(1-v^(2)//c^(2)) ● The mass of the spaceship increases by a factor sqrt(1-v^(2)//c^(2)) ● Mass and energy are interconvertable E = mc^(2) The speed of a meterial object can never exceed the velocity of light. ● If two objects A and B are moving with velocity u and v w.r.t each other along the x -axis, the relative velocity of A w.r.t. B = (u-v)/(1-uv//v^(2)) The momentum of an electron moving with a speed 0.6 c (Rest mass of electron is 9.1 xx 10^(-31 kg )

A physical quantity is a phyical property of a phenomenon , body, or substance , that can be quantified by measurement. The magnitude of the components of a vector are to be considered dimensionally distinct. For example , rather than an undifferentiated length unit L, we may represent length in the x direction as L_(x) , and so forth. This requirement status ultimately from the requirement that each component of a physically meaningful equation (scaler or vector) must be dimensionally consistent . As as example , suppose we wish to calculate the drift S of a swimmer crossing a river flowing with velocity V_(x) and of widht D and he is swimming in direction perpendicular to the river flow with velocity V_(y) relation to river, assuming no use of directed lengths, the quantities of interest are then V_(x),V_(y) both dimensioned as (L)/(T) , S the drift and D width of river both having dimension L. with these four quantities, we may conclude tha the equation for the drift S may be written : S prop V_(x)^(a)V_(y)^(b)D^(c) Or dimensionally L=((L)/(T))^(a+b)xx(L)^(c) from which we may deduce that a+b+c=1 and a+b=0, which leaves one of these exponents undetermined. If, however, we use directed length dimensions, then V_(x) will be dimensioned as (L_(x))/(T), V_(y) as (L_(y))/(T), S as L_(x)" and " D as L_(y) . The dimensional equation becomes : L_(x)=((L_(x))/(T))^(a) ((L_(y))/(T))^(b)(L_(y))^(c) and we may solve completely as a=1,b=-1 and c=1. The increase in deductive power gained by the use of directed length dimensions is apparent. From the concept of directed dimension what is the formula for a range (R) of a cannon ball when it is fired with vertical velocity component V_(y) and a horizontal velocity component V_(x) , assuming it is fired on a flat surface. [Range also depends upon acceleration due to gravity , g and k is numerical constant]

There is another useful system of units, besides the SI/mKs. A system, called the cgs (centimeter-gram-second) system. In this system Coloumb's law is given by vecF =(Qq)/(r^(2)).hatr where the distance is measured in cm (= 10^(-2) m), F in dynes (= 10^(-5) N) and the charges in electrostatic units (es units), where 1 es unit of charge 1/(3) xx 10^(-9) C . The number [3] actually arises from the speed of light in vacuum which is now taken to be exactly given by c = 2.99792458 xx 10^8 m/s. An approximate value of c then is c = [3] xx 10^8 m/s. (i) Show that the coloumb law in cgs units yields 1 esu of charge = 1 (dyne) 1/2 cm. Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L. (ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives: 1/(4piepsilon_(0)) = 10^(-9)/x^(2) (Nm^(2))/C^(2) with x=1/[3] xx 10^(-9) , we have 1/(4pi epsilon_(0)) = [3]^(2) xx 10^(9) (Nm^(2))/C^(2) or 1/(4pi epsilon_(0)) = (2.99792458)^(2) xx 10^(9) (Nm^(2))/C^(2) (exactly).