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uranmaximum [27]
2 years ago
8

A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from

the mass of a proton. If the HIHI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring?
Physics
1 answer:
vlada-n [284]2 years ago
0 0

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency f=7.0\times10^{13}\ Hz

We need to calculate the reduced mass

Using formula of reduced mass

\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}

Where, m_{H}= atomic mass of H

m_{I}= atomic mass of I

Put the value into the formula

\mu=\dfrac{1\times126.9}{1+126.9}

\mu=0.99\ u

\mu=0.99\times1.66\times10^{-27}\ Kg

\mu=1.643\times10^{-27}\ kg

We need to calculate the force constant of the spring

Using formula of frequency

f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}

k=f^2\times 4\pi^2\times\mu

Put the value into the formula

k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}

k=317.8\ N/m

Hence, The force constant of the spring is 317.8 N/m.

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olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground =  46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

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here d is distance and g is  acceleration and t is time

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h = 4.9 t²   ....................2

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0.54 h = 4.9 (t-1)²       ...................3

so

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0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

we use here 3.771 s because  0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

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h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

so height is 69.68 m

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2 years ago
A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency
allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

\nu=\dfrac{v}{\lambda}

\nu=\dfrac{251\ m/s}{0.051\ m}

\nu=4921.56\ Hz

(2) Angular frequency of the wave is given by :

\omega=2\pi\nu

\omega=2\pi\times 4921.56\ Hz

\omega=30923.07\ rad/s

(3) The period of oscillation is given by T as :

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T=\dfrac{1}{4921.56}

T = 0.000203 seconds

or

T = 0.203 milliseconds

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If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
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The formula that is applicable from physics is 
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