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3 years ago

How can you make the moon go around in a bigger circle

Physics

2 answers:

zlopas [31]3 years ago

5 0

You would have to give it more mechanical energy.

Like, strap a bunch of powerful rockets to one side of the moon, with all of them pointing in the direction that the moon is already moving in its orbit. Then blast away.

NOTE: There aren't enough rockets or rocket fuel on Earth to make a difference, even if you used ALL of them. The mass of the moon is about

<em>73,476,730,900,000,000,000,000 kilograms</em>

(rounded to the nearest hundred trillion kilograms.)

That's a lot.

Alex_Xolod [135]3 years ago

5 0

I doubt it if you can try

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The low areas created as a sound wave propagates are called rarefactions

qwelly [4]

The answer is; pressure

The sound is a longitudinal wave meaning the particles vibrate parallel to the direction of the wave. Sound waves, therefore, produce compression (akin to the crest in a transverse wave) and rarefaction regions (akin to a trough in a transverse wave) as its energy is propagated in the medium.

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3 years ago

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Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$

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3 years ago

When reading the printout from a laser printer, you are actually looking at an array of tiny dots.

solniwko [45]

Answer:

The value is $y = 3.097 * 10^{-5} \ m$

Explanation:

From the question we are told that

The diameter of the pupil is $d_p = 4.2 \ mm = 4.2 *10^{-3} \ m$

The distance of the page from the eye $d = 29 \ cm = 0.29 \ m$

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9} \ m$

The refractive index is $n_r = 1.36$

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

$y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d$

$y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29$

$y = 3.097 * 10^{-5} \ m$

7 0

3 years ago

In the first law of Thermodynamics ΔE = Q - W, what does ΔE stand for???

Alexxx [7]

<span>Δ</span>E = q + w

q = heat (quantity of)

q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system

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2 years ago

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A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl

raketka [301]

The wavelength of the note is $\lambda = 39.1 cm = 0.391 m$ . Since the speed of the wave is the speed of sound, $c=344 m/s$ , the frequency of the note is
$f= \frac{c}{\lambda}=879.8 Hz$

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where $\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m$ is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
$L= \frac{1}{2f} \sqrt{ \frac{T}{\mu} } =0.31 m$

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3 years ago