To find mean velocity use the following expression

 

Mean velocity = Total distance travelled/ total time taken

sin2A=cos3A

 

or sin2A=sin(/2 - 3A)

 

or 2A = (/2 - 3A)

 

or 5A = /2

 

or A =  /10

 

so, sinA = sin /10

where i denotes -1. This is an equation which allows you to interpret the exponentiation of an imaginary number ix as having a real part, cos x, and an imaginary part, i sin x. This was an especially useful observation in the solution of differential equations. Because of this and other uses of i, it became quite acceptable for use in mathematics. Euler, recommended the general use of these imaginary numbers

 

now,

e^-ix = cos x - i sin x          (2)

 From (1) and (2)

e^ix e^-ix = (cos x + i sin x )(cos x - i sin x )

 

or, 1 = cos²x - i² sin²x

or, 1 - cos²x  = - i² sin²x

or, sin²x = - i² sin²x

or,  i² = -1

or,   i = -1

The General Conference on Weights and Measures is the English name of the Conférence générale des poids et mesures (CGPM, never GCWM). It is one of the three organizations established to maintain the International System of Units (SI) under the terms of the Convention du Mètre (Metre Convention) of 1875. It meets in Paris every four to six years. In 2002 the CGPM represented 51 member states and ten further associate members.

Dear Akanksha

 

Remember following rules

 

1) The number of permutations of 'n' different thing takng 'r' at a time without repetition is

ⁿP_r = n!/(n-r)!

So if u go by this approach then in the question there are 4 different letters ( as A is repeated thrice) so ways in which word (containing four letters ) can be formed is

⁴P₄ = 4!/(4-4)! = 4!, what i have suggested in previous response to this question.

 

 

 

 

 

Consider an equilateral ABC with one of the vertex say A on the origin and side AB along X-axis. So coordinates of the vertices can be assigned as

A(0,0); B(a,0) and C(a/2, a3/2)

If masses m, 2m and 3m are placed on the vertices A, B and C respectively then coordinates of center of mass C(X,Y) can be expressed as

 

X = [0*m + a*2m + (a/2)*3m]/6m

 

or X= 7a/12

 

Similarly Y = [0*m + 0*2m + (a3/2)*3m]/6m = a3/4

 

so Y = a3/4

 

Now consider coordinates of square ABCD with side 'a' to be

A(0,0); B(0,a); C(a,a) and D(0,a) with masses m, 2m, 3m and 4m respectively and try to locate the center of mass.

Any rotating frame is Non-Inertial frame of reference, so Earth is ofcourse a non-inertial frame as it is rotating

 

Radius of earth = 6400km

Rotation of earth about its axis completes in 24hrs.

so angular displacement of earth in 24 hrs = 2 = 360 degrees

Angular speed is therefore =  = 360/(24*60) deg/min

                                               = 0.25 deg/min

 

Which is very small and thus we have strong reason to assume earth to be an inertial frame of reference.

 

This rotation of earth is however responsible for coriolis force that is maximum at equator and zero at poles.

This equation is definitely not a linear equation as the expansion of exponential term involve higher powers of variable r . Hence the equation may have multiple roots.

To solve this equation proceed as:

50r = e^20r  -  1

or 50r + 1 = e^20r

Take natural log of both the sides and proceed

 

 

A fictitious force is an apparent force that acts on all masses in a non-inertial frame of reference, e.g., a rotating reference frame. The force F does not arise from any physical interaction, but rather from the acceleration a of the non-inertial reference frame itself. Due to Newton's second law F = ma, fictitious forces are always proportional to the mass m being acted upon.

 

It is sometimes convenient to solve physical problems in a non-inertial reference frame. In such cases, it is necessary to introduce fictitious forces to account for the acceleration of the reference frame. For example, the surface of the Earth is a rotating reference frame. To solve classical mechanics problems exactly in an Earth-bound reference frame, two fictitious forces must be introduced, the Coriolis force and the centrifugal force (described below), of which the Coriolis force is dominant on Earth. Both of these fictitious forces are weak compared to most typical forces in everyday life, but they can be detected under careful conditions. For example, Léon Foucault was able to show the Coriolis force that results from the Earth's rotation using the Foucault pendulum. If the Earth were to rotate a thousand-fold faster (making each day only ~86 seconds long).

 

 

If a person living inside a closed box that is rotating (or otherwise accelerating) can detect their own rotation/acceleration. Careful observers within the box can detect that they are in a non-inertial reference frame from the fictitious forces that arise from the acceleration of the box. They can even map out the magnitude and direction of the acceleration at every point within the box. For example, a Foucault pendulum in a science museum will precess in exactly the same manner, regardless of whether the museum has walls or not.

For comparison, observers living inside a closed box that is moving uniformly (i.e., without acceleration) cannot detect their own motion. That is the essential physics of Newton's first two laws of motion.

 

 

 

When a car accelerates hard, the common human response is to feel "pushed back into the seat." In an inertial frame of reference attached to the road, there is no physical force moving the rider backward. However, in the rider's non-inertial reference frame attached to the accelerating car, there is a backward fictitious force. We mention two possible ways of analyzing the problem:

 

1. From the viewpoint of an inertial reference frame with constant velocity matching the initial motion of the car, the car is accelerating. In order for the passenger to stay inside the car, a force must be exerted on him. This force is exerted by the seat, which has started to move forward with the car and compressed against the passenger until it transmits the full force to keep the passenger inside. Thus the passenger is accelerating in this frame, due to the unbalanced force of the seat.
2. From the point of view of the interior of the car, an accelerating reference frame, there is a fictitious force pushing the passenger backwards, with magnitude equal to the mass of the passenger times the acceleration of the car. This force pushes the passenger back into the seat, until the seat compresses and provides an equal and opposite force. Thereafter, the passenger is stationary in this frame, because the fictitious force and the (real) force of the seat are balanced.

 

This serves as an illustration of the manner in which fictitious forces arise from switching to a non-inertial reference frame. Calculations of physical quantities made in any frame give the same answers, but in some cases calculations are easier to make in a non-inertial frame. (In this simple example, the calculations are equally easy in either of the two frames described.)

 

A similar effect occurs in circular motion, circular for the standpoint of an inertial frame of reference attached to the road, with the fictitious force called the centrifugal force, fictitious when seen from a non-inertial frame of reference. If a car is moving at constant speed around a circular section of road, the occupants will feel pushed outside, away from the center of the turn. Again the situation can be viewed from inertial or non-inertial frames:

 

 

To consider another example, taking as our reference frame the surface of the rotating earth, centrifugal force reduces the apparent force of gravity by about one part in a thousand, depending on latitude. This is zero at the poles, maximum at the equator.

 

Another fictitious force that arises in the case of circular motion is the Coriolis force, which is ordinarily visible only in very large-scale motion like the projectile motion of long-range guns or the circulation of the earth's atmosphere. Neglecting air resistance, an object dropped from a 50 m high tower at the equator will fall 7.7 mm eastward of the spot below where it was dropped because of the Coriolis force.

 

Both the centrifugal and the Coriolis force are needed to explain the motion of distant objects relative to rotating reference frames. Consider a distant star observed from a rotating spacecraft. In the reference frame co-rotating with the spacecraft the distant star appears to rotate around the spacecraft. The apparent motion of the star requires a fictitious centripetal force acting on the star. Just like in the example of the car in circular motion above, the centrifugal force acting on the star has the same magnitude as the centripetal force, but is directed in the opposite direction. In this case the Coriolis force has twice the magnitude of the centrifugal force and is directed oppositely to the centrifugal force.

 

Fictitious forces can be considered to do work, provided that they move an object on a trajectory that changes its energy from potential to kinetic. For example, consider a person in a rotating chair holding a weight in his outstretched arm. If he pulls his arm inward, from the perspective of his rotating reference frame he has done work against centrifugal force. If he now lets go of the weight, from his perspective it spontaneously flies outward, because centrifugal force has done work on the object, converting its potential energy into kinetic. From an inertial viewpoint, of course, the object flies away from him because it is suddenly allowed to move in a straight line. This illustrates that the work done, like the total potential and kinetic energy of an object, can be different in a non-inertial frame than an inertial one.

 

All fictitious forces are proportional to the mass of the object upon which they act, which is also true for gravity. This led Albert Einstein to wonder whether gravity was a fictitious force as well. He noted that a freefalling observer in a closed box would not be able to detect the force of gravity; hence, free falling reference frames are equivalent to an inertial reference frame (the equivalence principle).

 

Consider a particle with mass m and position vector x_a(t) in a particular inertial frame A. Consider a non-inertial frame B whose position relative to the inertial one is given by X(t). Since B is non-inertial, we must have that d²X/dt² (the acceleration of frame B with respect to frame A) is non-zero. Let the position of the particle in frame B be x_b(t). Then we have

Taking two time derivatives, this gives

Now consider the forces in the problem. By Newton's Second Law, F = ma. The true force is of course the one in frame A (the inertial one), so

However, suppose we are working to solve a problem in frame B. It may be useful to consider the apparent force in this frame, which is given by

Now we define

giving finally:

Thus we can solve problems in frame B by assuming that Newton's Second Law holds (with respect to quantities in that frame) and treating F_fictitious as an additional force.

 

A common situation in which noninertial reference frames are useful is when the reference frame is rotating. Since such rotational motion is non-inertial, due to the acceleration present in any rotational motion, a fictitious force can always be invoked by using a rotational frame of reference. Despite this complication, the use of fictitious forces often simplifies the calculations involved.

The relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with angular velocity can be expressed as

where we have used the relationship for the time derivative of a vector in rotating coordinates

Since , the acceleration becomes

or, equivalently,

The acceleration in the rotating frame equals

Since the force in the rotating frame is and, by definition, , the fictitious force equals

Here, the first term is the Coriolis force, the second term is the centrifugal force, and the third term is the Euler force. When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

 tan^(1/3)x dx

Let tanx=u^3

=>(sec x)^2 dx=3u^2 du

now, (sec x)^2=1+ (tan x)^2=1+u^6

=>dx=[(3u^2)/(1+ u^6)] du

therefore, ?tan^(1/3)x dx=3?[u X u^2/(1+u^6)]du

=>3?[u^3/(1+u^6)]du
let u^2=t
=>(2u) du=dt

=>3?[u^3/(1+u^6)]du=(3/2)?[t/(...

Now, partial fractions will be used
we know that, t/(1+t^3)=t/(1+t)(1+t^2-t)....... a^3 + b^3=(a+b)(a^2 + b^2 + ab)

let, t/(1+t)(1+t^2-t)=A/(1+t) + (Bt + C)/(1+t^2-t)

=>t/(1+t)(1+t^2-t)=[A(1+t^2 -t) + (Bt + C)(1+t)]/(1+t)(1+t^2 -t)................taking LCM

equating coefficients of u, u^2, constants on both sides
=>1=-A+B+C...................e... coefficients of u
0= A+B .......................equatin... coefficients of u^2
0= A+C .......................equatin... coefficients of constant
=>A=-1/3, B=1/3, C=1/3

=> (3/2)?[t/(1+t^3)]dt=
(3/2)?[(1/3)/(1+t)]dt + (3/2)?[(1/3)(1+t)/(1+t^2-t)]dt


=>(-1/2)log|1+t| + (1/2?[(1+t)/(1+t^2-t)]dt

=>(-1/2)log|1+t| + (1/4)?[(2+2t)/(1+t^2-t)]dt

we know, d/dt(1+ t^2 - t)=2t -1
=>(-1/2)log|1+t| + (1/4?[(2+2t+1-1)/(1+t^2-t)]dt

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| + (3/4)?1/(1+t^2-t)dt

we know, 1+t^2-t=(t-1/2)^2 +3/4

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| + (3/4)?1/[(t-1/2)^2 + 3/4]dt

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| +
(3/4)1/(2/sqrt(3) arctan[2(t-1/2)/sqrt(3)


substituting original values t=u^2

=>(-1/2)log|1+u^2| + (1/4)log|1+u^4 -u^2| +
(3/4)1/(2/sqrt(3) arctan[2(u^2-1/2)/sqrt(3)

substituting original values u=(tanx)^1/3

=>(-1/2)log|1+(tanx)^2/3| + (1/4)log|1+(tanx)^4/3 -(tanx)^2/3| +
(3/4)1/(2/sqrt(3) arctan[2((tanx)^2/3 - 1/2)/sqrt(3) + C

where arctan(x) means tan inverse x
results used=
1)?1/(1+x)dx= log|1+x| + c
2)?1/(a^2+x^2)dx=
(1/a)arctan(x/a) + C...........'a' is a constant

 

Now to work out for the problems based on relative density then the approach using an example is as follows:

 

Problem) A body floats with 40% of its volume above water. Find the relative density of the body.

 

Solution) Let Volume of the body = V

Density of the body = _b

Density of water = _w

Volume of the body inside the water = 0.6V

Hence, Buoyant force = 0.6V_wg

So, Buoyant force on the body due to water = Weight of the body

or  0.6V_wg = (V_b)g

or _b/_w= 0.6 which is relative density of the body.

Relative density (also known as specific gravity) is a measure of the density of a material. It is dimensionless, equal to the density of the material divided by some reference density (most often the density of water, but sometimes the air when comparing to gases):

Relative Density = _object / _reference

where  denotes density.

 

Since water's density is 1.0 × 10³ kg/m³ in SI units, the relative density of a material is approximately the density of the material measured in kg/m³ divided by 1000 (the density of water). There are no units of measurement.

Water's density can also be measured as nearly one gram per cubic centimeter (at maximum density) in non-SI units. The relative density therefore has nearly the same value as density of the material expressed in grams per cubic centimeter, but without any units of measurement.

Relative density or specific gravity is often an ambiguous term. This quantity is often stated for a certain temperature. Sometimes when this is done, it is a comparison of the density of the commodity being measured at that temperature, with the density of water at the same temperature. But they are also often compared to water at a different temperature.

 

 

Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale. For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic. But on the microscopic scale it is a seething mass of high speed molecules traveling at hundreds of meters per second.

 

Internal energy involves energy on the microscopic scale. For an ideal monoatomic gas, this is just the translational kinetic energy of the linear motion of the "hard sphere" type atoms , and the behavior of the system is well described by kinetic theory. However, for polyatomic gases there is rotational and vibrational kinetic energy as well. Then in liquids and solids there is potential energy associated with the intermolecular attractive forces. A simplified visualization of the contributions to internal energy can be helpful in understanding phase transitions and other phenomena which involve internal energy.

 

                             Internal Energy Example

 

When the sample of water and copper are both heated by 1°C, the addition to the kinetic energy is the same, since that is what temperature measures. But to achieve this increase for water, a much larger proportional energy must be added to the potential energy portion of the internal energy. So the total energy required to increase the temperature of the water is much larger, i.e., its specific heat is much larger.

The transfer of heat is normally from a high temperature object to a lower temperature object. Heat transfer changes the internal energy of both systems involved according to the First Law of Thermodynamics.

 

Consider a triangle ABC with centroid at G and the coordinaes be

 

A(0,0); B(x₁,y₁); C(x₂,y₂)

 

then coordinates of point 'G' the centroid of the triangle is given by

 

x = (0 + x₁ + x₂) /3 = (x₁ + x₂) /3

 

y = (0 + y₁ + y₂) /3 = (y₁ + y₂) /3

 

Now to prove

 

AB²+BC²+CA² = 3(GA²+GB²+GC²)

 

Substitute the coordinates at each sides and simplify to obtain above result.

 

Similarly take O(x,y) to prove second result