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butalik [34]
3 years ago
7

The focused, penetrating sound of the koto, which is similar to the tone of Japanese vocal music, comes from the_______________

a. Tension of the stringsc. Relaxation of strings d. Sound wave overlapping d. None of the above
Physics
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

a. Tension of the strings

Explanation:

The koto is a pluckedstring instrument. In these instruments, the sound is obtained causing the vibration in several strings. Therefore, its sound depends on the thickness, extension and tension of the strings. The lower the tension the deeper the sound will be.

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Coefficient of friction x mass x g
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3 years ago
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Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a
dusya [7]

Answer:

W=-21,870,000\ J

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

W=\Delta k=k_1-k_0

Since

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}

W=0-21870000\ J

\boxed{W=-21,870,000\ J}

3 0
3 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
We all have a tendency to make illusory correlations from time to time. Try to think of an illusory correlation that is held by
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Answer:

Thats a personal question

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It asks about you

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5 0
2 years ago
A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its
sdas [7]
<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

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We have equation of motion s = ut + 0.5 at²

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That is

           s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² -  u x 2 - 0.5 x g x 2²

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Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² -  u x 4 - 0.5 x g x 4²

           s₅ = u - 4.5 g    

That is

           u - 2.5 g = u - 4.5 g + 3

             2 g = 3

                g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

8 0
3 years ago
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