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3 years ago

In what different ways can be sound heard

Physics

2 answers:

Mekhanik [1.2K]3 years ago

5 0

The question is a bit ambiguous, but sound can be heard at different frequencies and amplitudes.

solong [7]3 years ago

4 0

-- through the pinna (the big scoopers that stick out on each side of our head that we call "ears")

-- through the mastoid ... the bone that bulges from the skull right behind our ears

-- through cochlear implants ... electrodes surgically implanted through the skull bones and into the inner ear

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a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance

8_murik_8 [283]

The car's (average) acceleration would be

$a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}$

The car's position over time would be given by

$x=v_0t+\dfrac12at^2$

so that after 2.4 seconds, the car will have traveled a distance of

$x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2$

$\implies x=77.5\,\mathrm m$

7 0

3 years ago

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The atomic mass is important because?

Jet001 [13]

B is correct answer.

The atomic mass is the most important because it tells you the number of electrons.

Hope it helped you.

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5 0

3 years ago

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A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N

Pepsi [2]

Answer:

The mass of the planet is $1.9407\times10^{27}\ kg$

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon $m=8.932\times10^{22}\ kg$

We need to calculate the mass of the planet

Using Kepler’s third law

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3$

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

$(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3$

$(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}$

$(8.932\times10^{22}+m)=1.94087\times10^{27}$

$m=1.94087\times10^{27}-8.932\times10^{22}$

$m=1.9407\times10^{27}\ kg$

Hence, The mass of the planet is $1.9407\times10^{27}\ kg$

8 0

4 years ago

Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be? Select all that apply.

lesya692 [45]

Answer:

option E

Explanation:

given,

Parallax angle(d) = 1 arcsecond

using Parallax formula

$d = \dfrac{1}{p}$

p is the parsecs angle which is measured in 1 arcsecond

d is the distance in parsec

now,

$P = \dfrac{1}{d}$

$P = \dfrac{1}{1}$

$P = 1 \ parsec$

we know,

1 parsec = 3.26 light year

hence, the answer will be option E

6 0

4 years ago

Please help

eimsori [14]

Answer:

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As the cold front passes, winds become gusty. There is a sudden drop in temperature, and also heavy rain, sometimes with hail, thunder, and lightning.

7 0

3 years ago

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In what different ways can be sound heard​

In what different ways can be sound heard