Fundamental frequency,
f=v2l=T/μ−−−−√2l
=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√
=58.96Hz
Let, n th harmonic is the hightest frequency, then
(58.93)n = 20000
∴N=339.38
Hence, 339 is the highest frequency.
∴fmax=(339)(58.93)Hz=19977Hz.
<h3>
What is frequency?</h3>
In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.
learn more about frequency refer:
brainly.com/question/254161
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The 'Bulge', the 'Disk', the 'Spiral Arms', and the 'Halo' those are four parts of a spiral galaxy. Try finding out the many different types of galaxies to understand which is which.
Answer:
The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.
Question:
Why did you lie about being in college?
Answer:it has risen
Explanation:
It has risen due to the burning of fossil fuels which produces a lot of green house gases.and it has also risen due to deforestation,no enough plants to consume the green house gas(carbon dioxide)
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
