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OleMash [197]
3 years ago
14

La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s

egundo es:
a-9.52 x 10-8 N
b-9.52 x 10-6 N
c-9.52 x 10-4 N
d-9.52 x 10-2 N
Physics
1 answer:
NemiM [27]3 years ago
8 0

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

For the waves in this string we have:

f=625 Hz, since it completes 625 cycles per second

\lambda=3.33 cm = 0.033 m is the wavelength

So the speed of the wave is

v=(625)(0.0333)=20.6 m/s

The speed of the waves in a string is related to the tension in the string by

v=\sqrt{\frac{T}{\mu}} (1)

where

T is the tension in the string

\mu=\frac{m}{L} is the linear density

In this problem:

m=16.5 g = 16.5\cdot 10^{-3} kg is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N

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<h2>Answer:</h2>

(a) 3.2 x 10²s

(b) 0.9 m/s (S 13 E)

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The sketch illustrating the scenario has been attached to this response.

As shown;

The fish swims due east with a velocity V_{x} = 0.2m/s

The river current has a velocity V_{y} due South = 0.9m/s

The resultant of the velocity is V

The width of the river is x = 64m

(a) To calculate how long it took the fish to get across the river, we know that velocity is the rate of change in distance, therefore we can use the relation;

V = \frac{d}{t}      -------------(i)

Where;

V = velocity of the fish = V_{x} = 0.2m/s

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t = time taken to move for that distance

Make t subject of the formula in equation (i);

t = \frac{d}{V}

Substitute the values of d and V into the equation;

t = \frac{64m}{0.2m/s}

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t = 3.20 x 10²s

Therefore, the time taken for the fish to get across the river is 3.20 x 10²s

(b) The resulting vector of the fish is V whose magnitude is the algebraic sum of vectors  V_{x} and  V_{y}, and direction is given by θ. i.e

<em>The magnitude of the resulting vector is;</em>

|V| = \sqrt{(V_x)^2 + (V_y)^2}

|V| = \sqrt{(0.2)^2 + (0.9)^2}

|V| = \sqrt{(0.04) + (0.81)}

|V| = \sqrt{(0.85)}

|V| = 0.92m/s

|V| ≅ 0.9m/s

<em>The direction of the resulting vector θ and is given by;</em>

tan θ = \frac{V_y}{V_x}

tan θ = \frac{0.9}{0.2}

tan θ = 4.5

θ = tan⁻¹ ( 4.5)

θ = 77.47° South of East.

θ  ≈ 77.5° South of East.

Subtracting θ = 77.5° from 90° gives its value East of South

i.e

90 - 77.5 = 12.5° East of South

<em>This can also be written as S12.5°E</em>

<em>Approximating to the nearest whole number gives </em>S 13 E

Therefore, the resulting velocity of the fish is 0.9m/s in the direction S13°E

(c) When the fish arrives on the opposite bank, its distance from being at the point directly across from where it started is the product of the velocity of the river current and the time taken by the fish to get across the river. This point is equivalent to k as shown in the diagram.

Therefore;

distance = velocity of river current x time taken

distance = 0.9m/s x 3.20 x 10²s

distance = 2.88 x 10²m

distance ≅ 2.9 x 10²m

<em>Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.</em>

<em />

<em />

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