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3 years ago

A tuba creates a 4th harmonic of

Physics

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

93.54 Hz

Explanation:

✓From the question, Number of harmonic frequency is 4

✓ the frequency (f₄ )= 116.5 Hz

✓harmonic frequency can be calculated using below expresion

fₙ = [ (nv)/4L]..........eqn(1)

v = speed of sound= 343 m/s

n = number of given harmonic frequency

L = Length of the rope

Using above expresion ,and substitute the values at (n=4) which is 4th harmonic frequency to find the " initial Lenght of the rope

fₙ = [ (nv)/4L]

f₄ = 4× 343 /4L

f₄ = 343 /L

L= 343 /f₄

But f₄= 116.5 Hz

L= 343/116.5= 2.944m

Hence, initial Lenght of the rope= 2.944m

We can determine the frequency of new length as ( initial Lenght of the rope + tubing Lenght)

= ( 2.944m + 0.721m )

= 3.667m

Hence, new length= 3.667m

To find the new frequency of the 4th harmonic we will use eqn(2)

f₄ = v/l ...............eqn(2)

From equation (2) If we substitute the values we have

f₄ = (343/3.667)

= 93.54 Hz

Hence, the the new frequency of the 4th harmonic is

93.54 Hz

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The work function of an element is the energy required to remove an electron from the surface of the solid. The work function fo

vichka [17]

Answer:

λ = 548.7 nm

Explanation:

Hi!

First we want to know how much energy we need to remove 1 electron from the surface of the solid:

218.1 kJ/mol => 218 100 J / (6.022 x 10^23) electrons

= 3.621 x 10^-19 J/electron

That is we need 3.621 x 10^-19 J to remove one electron

Now we can calculate the wavelength that a photon must have in order to have this energy:

E = (hc) / λ

λ = (hc) / Ε

where

h = 6,626070150(69) ×10 -34 Js (wikipedia)

c = 3 x10^8 m/s

hc = 1.987 x 10^-25 Jm

Therefore:

λ = ( 1.987 x 10^-25 /3.621 x 10^-19 ) m = 5.487 x 10^-7 m

λ = 548.7 nm

4 0

3 years ago

Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge

MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

$V_1 = V_2$ (1)

Keeping in mind the relationship between charge, voltage and capacitance:

$C=\frac{Q}{V}$

we can re-write (1) as

$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

$C=4 \pi \epsilon_0 R$

where R is the radius of the sphere. Substituting this into (2), we find

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}$ (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

$R_2 = 2R_1$

So the eq.(3) becomes

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}$

And re-arranging it we find:

$Q_2 = 2Q_1$

And since we know that the total charge is

$Q_1 + Q_2 = 7.00 nC$

we find

$Q_1 = 2.33 nC\\Q_2 = 4.67 nC$

3 0

4 years ago

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

bagirrra123 [75]

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship $v= 0.201c$

Wavelength of laser $\lambda= 697\ nm$

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

$\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}$

$\lambda=5.685\times10^{-7}\ m$

$\lambda=568.5\times10^{-9}\ m$

$\lambda=568.5\ nm$

Hence, The wavelength of observed light on earth is 568.5 nm

8 0

4 years ago

To understand the relationships between the parameters that characterize a wave. It is of fundamental importance in many areas o

Monica [59]

Answer:

1) c.

2) a.)

Explanation:

At any wave, if its waveform repeats itself every time interval T, it is said that the wave is periodic, with a period T, which is the time needed to complete an entire cycle. The other options refer at the way in the waves propagates (longitudinal or transversal) and to the type of waveform (sinusoidal), so the right answer is c).

At any wave that propagates at a constant speed, there exists a fixed relationship between the velocity v, the frequency f and the wavelength λ, as follows:

$v = \lambda * f (1)$

So in order to v keep constant, if the frequency is increased, the wavelength will decrease in the same proportion, so a) is the right answer.

7 0

3 years ago

A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down

Len [333]

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

4 0

3 years ago