Nodes and antinodes are part of a standing wave pattern.
The wavelength of the standing wave on a string that is 1.5 m long can be calculated using the formula: wavelength = 2L/n, where L is the length of the string and n is the number of nodes or antinodes.
A node is a point along a standing wave where the wave has minimal amplitude. The opposite of a node is an antinode, a point where the amplitude of the standing wave is a maximum. These occur midway between the nodes. Examples of a type of boundary could be the attachment point of a string, the closed end of an organ pipe or a woodwind pipe, the periphery of a drumhead, or a transmission line with the end short circuit. In this type, the amplitude of the wave is forced to zero at the boundary, so there is a node at the boundary, and the other nodes occur at multiples of half a wavelength from it: 0, λ/2, λ, 3λ/2, 2λ, ... One wavelength has 3 nodes and 2 antinodes. Between two nodes is the distance of half the wavelegth.
we know the equation V=nLv=speed of wave n=frequency L=wavelength here v=? n=250Hz & L=1.2 v=250*1.2 =300m/s
When a wave transfers into a denser medium, a portion of the wave is reflected back into the original medium. The reflected wave can have different amplitudes and phases compared to the incident wave depending on the properties of the medium interface.
The Fermi wave vector (k_F) of a fermi gas in a one-dimensional box of length L is given by the formula: k_F = n * π, where n is the number density of fermions in the box.
The wavelength of the standing wave on a string that is 1.5 m long can be calculated using the formula: wavelength = 2L/n, where L is the length of the string and n is the number of nodes or antinodes.
A node is a point along a standing wave where the wave has minimal amplitude. The opposite of a node is an antinode, a point where the amplitude of the standing wave is a maximum. These occur midway between the nodes. Examples of a type of boundary could be the attachment point of a string, the closed end of an organ pipe or a woodwind pipe, the periphery of a drumhead, or a transmission line with the end short circuit. In this type, the amplitude of the wave is forced to zero at the boundary, so there is a node at the boundary, and the other nodes occur at multiples of half a wavelength from it: 0, λ/2, λ, 3λ/2, 2λ, ... One wavelength has 3 nodes and 2 antinodes. Between two nodes is the distance of half the wavelegth.
With n nodes and b branches a network will have combination are
1014 it is. no of different trees possible with n nodes is (2^n)-n thanx
The maximum height of a binary tree with 'n' nodes is 'n-1'.
A binary tree with n nodes has exactly n+1 null nodes or Null Branches. so answer is 21. MOHAMMAD SAJID
let suppose total number of nodes/computers = n the formula will be = n(n-1)/2 e.g = 6(6-1)/2 =15 links.
If the number of levels is L, the maximum number of nodes N in a binary tree is N = 2L-1. For L = 5, N equates to 31 thus.
For an s orbital, there are no angular nodes. For a p orbital, there is 1 angular node. For a d orbital, there are 2 angular nodes. The maximum number of angular nodes is given by n-1, where n is the principal quantum number of the orbital.
As in one wavelength we have two anti-nodes so for 6 wavelengths we will have 12 anti-nodes.
For the height `h' of a binary tree, for which no further attributes are given than the number `n' of nodes, holds:ceil( ld n)
Level N of a binary tree has, at most, 2^N nodes. Note that the root node is regarded as being level 0. If we regard it as being level 1, then level N would have 2^(N-1) nodes at most.