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IRISSAK [1]
3 years ago
13

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

wtons. What is the frequency of this mode of vibration
Physics
1 answer:
Furkat [3]3 years ago
7 0

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

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Explanation:

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A golf ball bounces down a flight of steel stairs, striking several steps on the way down, but never hitting the edge of a step.
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Answer:

2.63m

Explanation:

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An atom's emission of light with a specific amount of energy confirms that
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How do you calculate the following:<br> Final Velocity?<br> Acceleration?<br> and Force?
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Wait like the equations, or is there an actual question?

Equations are

Final velocity (Vf) = Initial velocity (Vi) + Acceleration (a) x Time (t)
Acceleration (a) = (Final velocity [Vf] - initial velocity [Vi]) divided by Time (t)
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(Short version)

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How many seconds are required to deposit 0.110 grams of magnesium metal from a solution that contains Mg2 ions, if a current of
lakkis [162]

Answer: The time required to deposit such amount of Mg is 886secs.

Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.

He also stated that;

96500C(1Farday) of electricity is required to deposit 1 mole of any metal.

For Mg^2+ +2e- ==>Mg

193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)

Which implies;

19300C will liberate 24g

xCwill liberate 0.110g

Where x is the amount of electricity to deposit 0.110g

x = (19300 × 0.110)/24

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Recall that Q =It

Where Q is the quantity of electricity, I is the current and t is the time taken

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