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kipiarov [429]
2 years ago
9

S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr

ack (Fig. P15.75). The force constant of the spring is k, and the equilibrium length is l . Assume all portions of the spring oscillate in phase and the velocity of a segment of the spring of length d x is proportional to the distance x from the fixed end; that is, vx = (x / l) v . Also, notice that the mass of a segment of the spring is dm = (m /l) dx . Find (b) the period of oscillation.
Physics
1 answer:
Elina [12.6K]2 years ago
7 0

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

<h3>What is period of oscillation?</h3>

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

#SPJ4

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The diameter of a copper atom is approximately 2.28e-10 m. The massof one mole of copper is 64 grams. Assume that the atoms area
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1) Mass of one copper atom: 1.063\cdot 10^{-22} kg

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3) Mass of the cubical block: 992 kg

Explanation:

1)

We are told here that the mass of one mole of copper is

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We also know that the number of atoms inside 1 mole of substance is equal to Avogadro number:

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This means that N_A atoms of copper have a mass of M = 64 g. Therefore, we can find the mass of one copper atom by dividing the total mass by the avogadro number:

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2)

We are told that the diameter of a copper atom is

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The total mass of the cubical block of copper will be given by the mass of one atom of copper multiplied by the total number of atoms, so:

M= N^3 m

where:

N^3 = 9.33\cdot 10^{24} is the number of atoms in the cube

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M=(9.33\cdot 10^{24})(1.063\cdot 10^{-22})=992 kg

So, the mass of the cubical block is 992 kg.

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