Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Meiosis creates the gamete cells, because these cells are used in reproduction they only have 26 chromosomes two of these cells join together to make a full 46 chromosomes.
Hope this helps! :)
Explanation:
It is given that,
A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.
In 2 seconds, distance covered by the mass is 12 cm.
In 1 seconds, distance covered by the mass is 6 cm
So, in 16 seconds, distance covered by the mass is 96 cm
So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.
To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective

Replacing with our values


Therefore the focal length of th eobjective lenses is 27.75cm
Explanation:
Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.
Hence,
= 0. And, as
= 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.
Therefore,
Q = -(19000 J + 2000 J)
= -21000 J
or, |Q| = 21000 J
Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.