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Akimi4 [234]
3 years ago
6

A clarinet sounds as a closed pipe. if a clarinet sounds a note with a pitch of 375 hz, what are the frequencies of the lowest t

hree harmonics produced by this instrument?
Physics
1 answer:
mote1985 [20]3 years ago
8 0
So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.

Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.

So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
.
etc (: I hope this helps.
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Why is it important to use the correct number of significant digits when
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Answer:

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Explanation:

Scientists use significant figures to avoid claiming more accuracy in a calculation than they actually know.

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A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?
adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

\sum F = 0

F_b -W = 0

F_b = W

F_b = mg

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

\rho_w V_{displaced} g = mg

Remember the expression for which you can determine the relationship between mass, volume and density, in which

\rho = \frac{m}{V} \rightarrow m = V\rho

In this case the density would be that of the object, replacing

\rho_w V_{displaced} g = V\rho g

Since the displaced volume of water is 0.429 we will have to

\rho_w (0.429V) = V \rho

0.429\rho_w= \rho

The density of water under normal conditions is 1000kg / m ^ 3, so

0.429(1000) = \rho

\rho = 429kg/m^3

The density of the object is 429kg / m ^ 3

7 0
2 years ago
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
Un objeto, que se encuentra a nivel del suelo, es lanzado verticalmente hacia arriba con una velocidad de 160 km/hr. ¿Qué altura
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wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

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Explanation:

QA = 4 micro Coulomb

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A is origin, B is at 4 m and P is at 6 m .

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The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
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