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oee [108]
3 years ago
7

En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia

de la Onda.
Physics
1 answer:
sattari [20]3 years ago
7 0

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

  • Velocidad = 50 m/s
  • Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

Sustituyendo en la fórmula, tenemos;

Frequencia = \frac {50}{40}

<em>Frequencia = 1.25 Hz</em>

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3 years ago
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5 0
2 years ago
Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to
Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x (\frac{Rc}{Rc + Ri})

where

  • Rc = resistance of copper = \frac{ρl}{a}  (l = length , a = area, ρ = resistivity of copper)
  • Ri = resistance of iron = \frac{ρ₀l}{a}  (l = length , a = area, ρ₀ = resistivity of copper)

Vc =  V x (\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}})

Vc = V x (\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})})

Vc = V x (\frac{ρ}{ρ + ρ₀})

where

  • ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
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Vc = 16 x (\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}})

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5 0
3 years ago
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so
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Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

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3 years ago
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