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1 year ago

If one oscillation has 5. 8 times the energy of a second one of equal frequency and mass, what is the ratio of their amplitudes?

Physics

1 answer:

quester [9]1 year ago

7 0

The ratio of their amplitudes will be 2.41.

To find the answer, we have to know more about the simple harmonic motion.

<h3>How to find the ratio of their amplitudes?</h3>

It is given that,

$E_1=5.8E_2\\m_1=m_2 \\\omega_1=\omega_2$

We have the relation between these quantities and amplitudes as,

$A=\sqrt\frac{2E}{k} \\k=m\omega^2$

Here, for both the oscillation, k will be same.

Thus, the ratio of amplitude will be,

$\frac{A_1}{A_2} =\sqrt{\frac{E_1}{E_2} } =\sqrt{\frac{5.8E_2}{E_2} } =2.41$

Thus, we can conclude that, the ratio of their amplitudes will be 2.41.

Learn more about simple harmonic motion here:

brainly.com/question/22422926

#SPJ4

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A rock, initially at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth.

lianna [129]

Answer:

the rock speed is increased

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3 years ago

A student who was training for a cross country race jogged for 2. 0 hours and covered a distance of 14. 0 kilometers. What was t

FinnZ [79.3K]

The average speed is 116.66m/s.

Given - The path traced is 14km , time for jogging is 2 hrs=120min

To find the average speed-

Speed refers to the ease of the movement and degree of mobility as a result of force application.
Due to this there is involvement of velocity.
Journey of average speed is the cumulative of distances and time.
Kinetic theory refers to the Boltzmann constant connecting to the standards of distance traversed.

calculations-

$Speed= \frac{Distance}{Time}$

Speed= 14 000 / 120

= 116.66m/s

To learn more about average speed -

<u>brainly.com/question/27753148</u>

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6 0

1 year ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

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