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MrRa [10]
2 years ago
10

The speed of light inside a medium is (2.0 x 10^8m/s) What is the index of refraction (n) of the medium?

Physics
1 answer:
kondor19780726 [428]2 years ago
3 0

Answer:

Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium

1.5 = 3 x 108 / Speed of light in medium

Speed of light in the medium = 3 x 108 /1.5

= 2 x 108 m/s.

Explanation:

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It goes triassic, Jurassic, Cretaceous
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Explanation:

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7 0
2 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
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5 0
3 years ago
A wire 6.60 m long with diameter of 2.05 mm has a resistance of 0.0310 Ω.
Alex73 [517]

Answer:

1.551×10^-8 Ωm

Explanation:

Resistivity of a material is expressed as shown;.

Resistivity = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the wire.

Given;

R = 0.0310 Ω

A = πd²/4

A = π(2.05×10^-3)²/4

A = 0.000013204255/4

A = 0.00000330106375

A = 3.30×10^-6m

l = 6.60m

Substituting this values into the formula for calculating resistivity.

rho = 0.0310× 3.30×10^-6/6.60

rho = 1.023×10^-7/6.60

rho = 1.551×10^-8 Ωm

Hence the resistivity of the material is 1.551×10^-8 Ωm

6 0
3 years ago
PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball
bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

             q₁  = 3C

              q₂ = 14C

Distance between balls  = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

                      F  = \frac{k q_{1} q_{2} }{r^{2} }

where k  = 9 x 10⁹Nm²/C²

           q is the charges

             r is the distance

Input the variables and solve;

                 

        F  = \frac{9 x 10^{9} x 3 x 14 }{9000}  = 4.2 x 10⁷N

8 0
3 years ago
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