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umka21 [38]
1 year ago
10

The difference in energy between allowed oscillator states in HBr molecules is 0.330 eV. What is the oscillation frequency of th

is molecule
Physics
1 answer:
SpyIntel [72]1 year ago
3 0

The oscillation frequency of this molecule is 7.96•10¹³ Hz.

<h3>What is Oscillation?</h3>

Oscillation is the repeating or periodic change of a quantity around a central value or between two or more states, often in time. Alternating current and a swinging pendulum are two common examples of oscillation.

There are 3 main types of Oscillation –

  • Free
  • damped
  • forced oscillation

ΔE=hf

f= ΔE/h=0.33•1.6•10⁻¹⁹/6.63•10⁻³⁴ = =7.96•10¹³ Hz

The oscillation frequency of this molecule is 7.96•10¹³ Hz.

to learn more about oscillation go to - brainly.com/question/12622728

#SPJ4

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Horace has invented a unique pair of reading glasses that have two small light bulbs at the bottom wired in series, so that he c
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Answer:

The current drawn by Horace’s reading glasses is 0.8 A.

Explanation:  

Given that,

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Voltage of the system, V = 3.2 volts

These two bulbs are connected in series. The equivalent resistance will be 2 ohms +2 ohms = 4 ohms

Let I is the current drawn by Horace’s reading glasses. Using Ohm's law to find it such that :

V=IR\\\\I=\dfrac{V}{R}\\\\I=\dfrac{3.2}{4}\\\\I=0.8\ A

So, the current drawn by Horace’s reading glasses is 0.8 A.

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3 years ago
The magnetic field 0.02 m from a wire is 0.1 t. what is the magnitude of the magnetic field 0.01 m from the same wire?
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3 years ago
Read 2 more answers
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val
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g Generally the accepted value of acceleration due to gravity is 9.801 m/s^2

as per the question the acceleration due to gravity is found to be 9.42m/s^2 in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is -\frac{0.381}{9.801} *100

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the error is not so high,so it can be  accepted.

now we have to know why this occurs-the equation of time period of the simple pendulum is give as-T=2\pi\sqrt[2]{l/g}

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As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801  m/s^2

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