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.
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Hey mate
Here is your answer
Option A)
Explanation:
The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.
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The most common liquid on planet earth is water
Answer:
The impulse is -20kg m/s.
Explanation:
given mass of ball = 0.4 kg
The ball is moving towards left with the initial velocity = 30m/s
The momentum in left direction P1 = mv1
= (0.4)(30)
= 12 kg m/s
The given velocity towards right = -20 m/s ( negative sign shows inverse direction)
Momentum towards right, P2 = mv2
P2 = (0.4)(-20)
P2 = -8 kg m/s
The Impulse = change in momentum
= P2 –P1
= -8-12
= -20 kg m/s
Answer:
C.O.P = 1.49
W = 335.57 joules
Explanation:
C.O.P = coefficient of performance = (benefit/cost) = Qc/W ...equ 1 where C.O.P is coefficient of performance, Qc is heat from cold reservoir, w is work done on refrigerator.
Qh = Qc + W...equ 2
W = Qh - Qc ...equ 3 where What is heat entering hot reservoir.
Substituting for W in equ 1
Qh/(Qh - Qc) = 1/((Qh /Qc) -1) ..equ 4
Since the second law states that entropy dumped into hot reservoir must be already as much as entropy absorbed from cold reservoir which gives us
(Qh/Th)>= (Qc/Tc)..equ 5
Cross multiple equ 5 to get
(Qh/Qc) = (Th/Tc)...equ 6
Sub equ 6 into equation 4
C.O.P = 1/((Th/Tc) -1)...equ7
Where Th is temp of hot reservoir = 493k and Tc is temp of cold reservoir = 295k
C.O.P = 1/((493/295) - 1)
C.O.P = 1.49
To solve for W= work done on every cycle
We substitute C.O.P into equ 1
Where Qc = 500 joules
1.49 = 500/W
W = 500/1.49
W = 335.57 joules
