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aniked [119]
3 years ago
10

The amplitude of a sound wave determines it's

Physics
2 answers:
Ket [755]3 years ago
7 0
Amplitude determines loudness
Relation between them is,
Loudness = 10㏒(I1/I2)

Where I1 and I2 are the intensities of 2 sound waves.
Snezhnost [94]3 years ago
5 0
To my best knowledge the answer is D- Wavelength 
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What is the number of significant digits in the measurement 34.000 m?
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A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
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