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tiny-mole [99]
3 years ago
12

What is the speed of light in water?

Physics
1 answer:
Fudgin [204]3 years ago
8 0

Answer:

Speed of the light in water= 225,000 km/s

Explanation:

At the speed with which light propagates through a homogeneous and transparent medium, it is a constant characteristic of that medium, and therefore, it changes from one medium to another.

Due to its enormous magnitude, the measurement of the speed of light has required the invention of ingenious procedures that will overcome the inconvenience of short land distances in relation to such extraordinary speed.

Astronomical methods and terrestrial methods have been giving ever closer results. At present, the value c = 299,792,458 km / s is accepted for the speed of light in a vacuum. In any transparent material medium the light propagates with a speed that is always lower than c. Thus, for example, in water it does so at around 75% of the speed of light in a vacuum: about 225,000 km / s.

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Two particles are separated by a certain distance. The force of gravitational interaction between them is F0. Now the separation
inysia [295]

Answer:

F'=1/9*F0

Explanation:

F0 is the gravitational force between the particles. When the distance is triplicated we have that

F'=G\frac{m_1m_2}{(3r)^{2}}

where r is the distance before the particles are separated, m1 and m2 are the masses their masses and G is the Canvendish's constant.

By some algebra we have

F'=\frac{1}{9}G\frac{m_1m_2}{r^2}=\frac{1}{9}F_0

hope this helps!!

4 0
4 years ago
Read 2 more answers
Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm50.0 cm long with cross-sectional area 1.00 mm21.00
deff fn [24]

Answer:

c < a = b

Explanation:

The tensile stress = Force applied/(Cross sectional area)

(a) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(b) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(c) The applied force = 100 N

The cross sectional area = 2.00 mm² = 2 × 10⁻⁶ m²

The tensile stress = 100 N/(2 × 10⁻⁶ m²) = 50,000,000 Pa = 50 MPa

Therefore, the tensile stress from smallest to largest are;

(a) 50 MPa, < (b) 200 MPa = (a) 200 MPa

Therefore, we have;

c < a = b.

8 0
3 years ago
What happens to the pressure of a gas when the volume of its container is increased at constant temperature?
zepelin [54]
<h2><em><u>Pressure </u></em><em><u>decreases </u></em><em>as </em><em>volume </em><em>increases </em><em>at </em><em>constant </em><em>temperature</em><em>.</em></h2>

<em>Hope </em><em>it</em><em> helps</em><em>.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

3 0
3 years ago
A 85 kg skier races down a hill from a height of 100 m. Ignore friction.
Ede4ka [16]

Answer:

(a)158.9 N up the ski slope

(b) 0.19

Explanation:

PLEASE MARK ME AS BRAINLIEST

F^k=MkN

158.9=817.7 M*s

Mk=0.19

4 0
3 years ago
The distance to the other planets in the Solar System from the Earth depends on where they are in their orbit. The closest that
Afina-wow [57]

Answer:

3.6 arcsec

Explanation:

 angular diameter = diameter / distance

diameter is constant

so angular diameter ∝ 1 / distance

angular diameter = k / distance

For first case ,

18 = k / .5

for second case let angular diameter be D .

D = k / 2.5

dividing ,

D / 18 = .5 / 2.5 = 1 / 5

D = 18 / 5 = 3.6 arcsec

3.6 arcsec is the answer .

3 0
3 years ago
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