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3 years ago

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A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

, what is the fundamental frequency

1 answer:

Gre4nikov [31]3 years ago

5 0

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

$V = \sqrt{\frac{T}{\mu} }$ where T is the tension in the string and $\mu$ is the density of the string

Given T = 600N and $\mu$ = 0.015 g/cm = 0.0015kg/m

$V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s$

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

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2 years ago

Read 2 more answers

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

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The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

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Answer:

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2 years ago

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Answer:

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Explanation:

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