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2 years ago

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Physics

1 answer:

slega [8]2 years ago

7 0

Answer:A language is a structured system of communication. Language, in a broader sense, is the method of communication that involves the use of – particularly human – languages. The scientific study of language is called linguistics

Explanation:

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A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

Umnica [9.8K]

Answer:

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

$\displaystyle w=\sqrt{\frac{k}{m}}$

The equation for the motion of the object is written as a sinusoid:

$\displaystyle X=A\ cos\ w\ t$

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

$\displaystyle X'=V=-A\ w\ sin\ w\ t$

And the maximum speed is

$\displaystyle V_{max}= A\ w$

Solving for the amplitude

$\displaystyle A= \frac{V_{max}}{w}$

Computing w

$\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s$

Calculating A

$\displaystyle A=\frac{1.7}{24.5}=0.069\ m$

$\displaystyle \boxed{A=6.9\ cm}$

7 0

3 years ago

During intercourse, Lydia constantly worries whether her partner thinks she is

Misha Larkins [42]

Answer is c spectatoring

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2 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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2 years ago

What does the voltage-current graph above show about the relationship between voltage and current, and how do these properties a

zubka84 [21]

The answer is as voltage increases current increases and therefore resistance would remain constant

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2 years ago

Write a hypothesis why the moon has very little liquid water.

Llana [10]

Because the Moon has a very small surface area compared to other spacial geo-bodies, it has cooled down much faster than Earth. Any water on the moon would freeze.

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3 years ago

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