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ELEN [110]
4 years ago
9

2. El sonido de una ballena es en especial de frecuencia baja, pero existe una especie de ballena la Whalien cuya frecuencia es

de 52 Hz, si la velocidad de propagación de la onda es de 1400m/s ¿Cuál será su periodo en el agua y el aire? Y ¿Cuál será la longitud de onda en cada medio? Recuerda que la velocidad de propagación en el aire es 340m/s
Physics
1 answer:
ikadub [295]4 years ago
8 0

Answer:

The sound of a whale is especially low frequency, but there is a species of whale the Whalien whose frequency is 52 Hz, if the propagation speed of the wave is 1400m / s What will be its period in the water and the air? And what will be the wavelength in each medium? Remember that the propagation speed in air is 340m / s

Explanation:

From wave equation, the speed, wavelength and frequency is related using

V = fλ

Where

V is the speed

f is the frequency

And λ is the wavelength

So,

The frequency of the whale is

f = 52Hz

The speed in water is V_w = 1400m/s

The speed in air is V_a = 340m/s

We want to find the period in each medium, the period is related to the frequency and since the frequency is constant.

Then, period in equal in both medium

T = 1 / f

T_w = T_a = 1 / f

T = 1 / 52

T = 0.0192 seconds

We want to find the wavelength in each medium

For water,

V = fλ

V_w = f × λ_w

Then,

λ_w = V_w / f.

λ_w = 1400 / 52 = 26.92 m

The wavelength in water is 26.92m

Now, in air

V = fλ

V_a = f × λ_a

Then,

λ_a = V_a / f.

λ_a = 340 / 52 = 6.54 m

The wavelength in air is 6.54 m

In Spanish

De la ecuación de onda, la velocidad, la longitud de onda y la frecuencia se relacionan usando

V = fλ

Dónde

V es la velocidad

f es la frecuencia

Y λ es la longitud de onda

Entonces,

La frecuencia de la ballena es

f = 52Hz

La velocidad en el agua es V_w = 1400m / s

La velocidad en el aire es V_a = 340m / s

Queremos encontrar el período en cada medio, el período está relacionado con la frecuencia y dado que la frecuencia es constante.

Luego, período igual en ambos medios

T = 1 / f

T_w = T_a = 1 / f

T = 1/52

T = 0.0192 segundos

Queremos encontrar la longitud de onda en cada medio

Para agua,

V = fλ

V_w = f × λ_w

Entonces,

λ_w = V_w / f.

λ_w = 1400/52 = 26,92 m

La longitud de onda en el agua es de 26,92 m.

Ahora en el aire

V = fλ

V_a = f × λ_a

Entonces,

λ_a = V_a / f.

λ_a = 340/52 = 6,54 m

La longitud de onda en el aire es de 6.54 m.

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son4ous [18]

Answer:

I think number 3

Explanation:

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Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g,
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Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\

Now equate the forces,

T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree

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A projectile is fired from ground level with an initial speed of 650 m/sec and an angle of elevation of 30 degrees. Use that the
svlad2 [7]

Answer:

(a) The range of the projectile is 37,336.3 m

(b) The maximum height of the projectile is 5,389.03 m

Explanation:

Given;

initial velocity of the projectile, u = 650 m/s

angle of projection, θ = 30⁰

(a) The range of the projectile is calculated as;

R = \frac{u^2sin 2\theta}{g} \\\\R = \frac{(650^2)Sin (2\times 30^0)}{9.8} \\\\R = 37,336.3 \ m

(b) The maximum height of the projectile is calculated as;

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3 years ago
A 60.0-kg person rides in an elevator while standing on a scale. the scale reads 400 n. the acceleration of the elevator is clos
Keith_Richards [23]

Reading of the scale will be given as Normal force on the man

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vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

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