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3 years ago

An explosion occurs 34 km away. The time it takes for its sound to reach your ears, traveling at 343 m/s, is

Physics

2 answers:

madam [21]3 years ago

7 0

Answer: 99.1 seconds

Explanation: You know that the total distance is 34km, and the speed of the sound is 343m/s.

Now, the total time that takes to something with a given velocity, to reach a distance D, is:

T = D/v

where D = 34km and v = 343m/s

Now, here we have different units, so we must convert the kilomeres into meters, this can be done as: 1km = 1000m then, 34km = 1000*34m = 34000m

Then the total time is:

T = 34000m/343m/s = 99.1 seconds

Ostrovityanka [42]3 years ago

6 0

Answer:

99 s

Explanation:

distance = rate × time

34000 m = (343 m/s) t

t = 99 s

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The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively, and their centers are separated by 3.84×108m.

bija089 [108]

Answer:

C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.

Explanation:

At first we assume that both Earth and Moon can be treated as particles, the center of mass of the Earth-Moon system is obtained by using this formula:

$r_{CM} = \frac{r_{E}\cdot m_{E}+r_{M}\cdot m_{M}}{m_{E}+m_{M}}$ (Eq. 1)

Where:

$r_{E}$ - Location of the center of the Earth, measured in kilometers.

$r_{M}$ - Location of the Moon, measured in kilometers.

$r_{CM}$ - Location of the center of mass, measured in kilometers.

$m_{E}$ - Mass of the Earth, measured in kilograms.

$m_{M}$ - Mass of the Moon, measured in kilograms.

If we know that $r_{E} = 0\,km$ , $r_{M} = 3.84\times 10^{8}\,m$ , $m_{E} = 5.98\times 10^{24}\,kg$ and $m_{M} = 7.35\times 10^{22}\,kg$ , the location of the center of mass respect to the Earth is:

$r_{CM} = \frac{(0\,km)\cdot (5.98\times 10^{22}\,kg)+(3.84\times 10^{8}\,m)\cdot (7.35\times 10^{22}\,kg)}{5.98\times 10^{24}\,kg+7.35\times 10^{22}\,kg}$

$r_{CM} = 4.662\times 10^{6}\,m$

The Earth has a radius of $6.371\times 10^{6}$ meters, we notice that center of mass in located inside the Earth and the radius of rotation of the Earth around the center of mass is much greater than the radius of rotation of the Moon around the center of mass. That center of mass follows an orbit around the sun.

In consequence, correct answer is C.

4 0

3 years ago

Help with these questions

erma4kov [3.2K]

<h2>Explanation:</h2><h3>3. </h3>

When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:

Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.

_____

The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...

n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33

none of the offered choices is correct. The closest is 1.34.

8 0

3 years ago

What are the SI units of thermal conductivity?

kotykmax [81]

Answer:

The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).

Explanation:

hope this will help u

7 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

3 years ago

Which phrase best describes the outer planets?

pav-90 [236]

Answer:

Astronomers have divided the eight planets of our solar system into the inner planets and the outer planets. The 4 inner planets are the closest to the Sun, and the outer planets are the other four – Jupiter, Saturn, Uranus, and Neptune. The outer planets are also called the Jovian planets or gas giants.

Explanation:

6 0

3 years ago

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