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iren2701 [21]
2 years ago
12

What machine would benefit from slowing lights speed

Physics
1 answer:
VARVARA [1.3K]2 years ago
8 0

Answer: exotic medium can be engineered to slow light

Explanation:

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write an essay about "the importance of knowing the fluids" / the importance of fluid knowledge dunno exactly. Plz help
Gala2k [10]
DRINK ALOT  of fluid is the anwser
6 0
3 years ago
Suppose that you are standing on a train accelerating at 0.20g (where g is the acceleration due to gravity). What minimum coeffi
prisoha [69]

Answer:

0.2

Explanation:

The given parameters are;

The acceleration of the train, a = 0.2·g

The mass of the person standing on the train = m

Let μ represent the coefficient of static friction, we have;

The force acting on the person, F = m × a = m × 0.2·g

The force of friction acting between the feet and the floor, F_f = m·g·μ

For the person not to slide we have;

The force acting on the person = The force of friction acting between the feet and the floor

F = F_f

∴ m × 0.2·g = m·g·μ

From which we get;

0.2 = μ

The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.

7 0
3 years ago
Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

6 0
3 years ago
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Alexxx [7]

Answer:

the answer to this question is 2,4,3,1

5 0
3 years ago
An object is dropped from a height of 25 meters. At what velocity will it hit the ground? A. 7.0 meters/second B. 11 meters/seco
Basile [38]
Final^{2}=Initial^{2}+2ad \\ x^{2}=0^{2}+2(9.8)(25) \\ x^{2}=490 \\ x=22.13 \\ C
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