Bryophyte pteridophyte Gymnosperm

Gravitational force at the midpoint will be zero. This is because both spheres are identical and their forces will be equal but opposite in direction.
The gravitational potential is given by :
or ...

Total energy = kinetic energy (KE) + potential energy(PE)
KE > 0 since m and v2 is positive. If KE <0 particles cannot be find. If PE>TE, then KE<0 (now in all graph check for this condition)
In case 1 kinetic energy is negative for x<a. So at x<a particle cannot be found.
In case 2 for x<a and for x> b kinetic energy is negative. So the particle cannot be found in these regions.
In the...

It cannot be said from the above-given graph whether the particle is moving along a straight line or along a parabolic path as the x-t graph does not tell us about the trajectory taken by the particle. From the given graph we can only say that the position of the particle along the x-axis does not change till time t=0 and after that it starts increasing in non-linear manner.

(i) (a) All the points vibrate with the same frequency of 60 Hz.
(b) They all have the same phase as it depends upon time.
(c) At different points, the amplitude is different and is equal to A(x) given by
(ii)

Let the equation of oscillation be given by
Velocity would be given as
Kinetic energy at an instant is given by
Time Period is given by
The Average Kinetic Energy would be given as follows
The potential energy at an instant T is given by
The Average Potential Energy would be given by
We can see Kav = Uav

As we know that the general term in the binomial expansion of is given by
So,
Term in the expansion of :
Term in the expansion of :
Term in the expansion of :
Now, As given in the question,
From here, we get ,
Which can be written as
From these equations we get,

Let A be a vector such that:-
Then the magnitude of vector A is given by :
Now let us assume that the angle made between vector A and x-axis is .
Then we have:-
Similarly, let B be a vector such that:-
The magnitude of vector B is :
Let be the angle...

As we know that if a point moves in a plane in such a way that its distance from two-point remain constant then the path is an ellipse.
Now, According to the question,
the distance between the point from where the sum of the distance from a point is constant = 10
Now, the distance between the foci=8
Now, As we know the relation,
Hence the equation of the ellipse is,
Hence the path of...

Given quantum numbers :
It is possible and it is 3p orbital.

Given quantum numbers :
It is possible and it is 2p orbital.

Given quantum numbers :
It is possible and it is 1s orbital.

Given quantum numbers :
NOT possible, because n cannot be equal to zero.

Total numbers of numbers in the draw = 20
Numbers to be selected = 6
n(S) =
Let E be the event that six numbers match with the six numbers fixed by the lottery committee.
n(E) = 1 (Since only one prize to be won.)
Probability of winning =

Sample space when three coins are tossed: [Same as a coin tossed thrice!]
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]
Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}
n(E) = 3
The required probability of getting exactly 2 tails is .

Given problem is
Now, we will reduce it into
Now, multiply numerator an denominator by
Therefore, answer is

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