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3 years ago

5

You are listening to music in your room and you are asked to turn down the volume of your music. When you turn down the volume w

hat will happen to the sound waves' amplitude?

1 answer:

djyliett [7]3 years ago

6 0

The sound waves will get smaller and smaller till there is no more bc it is off.

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A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. In

Sveta_85 [38]

Answer:

$E_M=0.0492J$ .

Explanation:

The mechanical energy of the system will be the kinetic energy plus the elastic potential energy: $E_M=K+U_e$ .

We know that the equation for the kinetic energy is $K=\frac{mv^2}{2}$ , where <em>m </em>is the mass of the object and <em>v </em>its velocity.

We know that the equation for the elastic potential energy is $U_e=\frac{k\Delta x^2}{2}$ , where <em>k</em> is the spring constant and $\Delta x$ the compression (or elongation) respect to equilibrium.

So for our values we have:

$E_M=K+U_e=\frac{mv^2}{2}+\frac{k \Delta x^2}{2}=\frac{(0.24kg)(0.4m/s)^2}{2}+\frac{(6N/m)(0.1m)^2}{2}=0.0492J$ .

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3 years ago

How does the government shape technology?

Marat540 [252]

Answer:

Society is a group of people that share similar values and beliefs.

The development of technology is affected by society and its changing values, politics, and economics

Explanation:

8 0

3 years ago

A satellite of 2,140 kg travels around the earth, of 5.97 - 1024

77julia77 [94]

Answer:

8,961

Explanation:

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3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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3 years ago

Finderkeepers,ismadeormodel

ankoles [38]

Answer:

model lol

Explanation:

7 0

3 years ago