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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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LenaWriter [7]

Answer: 7.41 m/s

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Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

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Answer:

The law of inertia

Explanation:

A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force

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3 years ago
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A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

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indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

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let's calculate

         v₂ = √( 2 9.8 250)

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weqwewe [10]

Answer:

41

Explanation:

and your welcome

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The correct answer is A. 250N

Work is a product of force and distance.

That is, work done=force×distance

Therefore substituting for the values in the question:

500J=force×2m

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another unit for work done is Nm as force as the SI unit of force is newtons and distance in meters

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3 years ago
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