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Elden [556K]
3 years ago
10

The speed of light in a material is 0.50

Physics
1 answer:
vivado [14]3 years ago
4 0

Answer:

Explanation:

This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum

Let's find the refractive index of the material

            n = c / v

           n = 3 10⁸ / 0.5 10⁸

           n = 6

When the material passes from one medium to another, it must comply with the law of refraction

           n₁ sin θ = n₂ sin θ₂

for the angle criticize the angles tea2 = 90

          tea = sin⁻¹n₂/ n₁

The vacuum replacement index is n₂ = 1

         tea = sin⁻¹ (1/6)

         tea = 9.59º

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two electronics students are discussing electrical units. student A says that electrical power is measured in units coulombs, st
Savatey [412]
The statements of both students are incorrect.

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-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms'  means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).

There are no 'following statements'.

All in all, a very disappointing question.

8 0
3 years ago
A car driving at a constant speed of 64 mi/h travels 68 miles. How many hours did this take?
babymother [125]
64 miles/hour
Therefore 1/64 hours/mile
68 miles * 1/64 hours/mile (notice how miles cancels out)
Therefore the answer is 68/64 hours = 1.0625 hours = 1 hour 3min and 45sec.
3 0
3 years ago
What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?
blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

7 0
3 years ago
The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu
UNO [17]

Answer:

=170kcal

Explanation:

We first calculate the amount of energy required to melt the alcohol using the formula: MLf, where Lf is the latent heat of fussion

We then calculate amount of heat required to raise the temperature of liquid alcohol to -14° C using MC∅.  We then add the two.

Thus ΔH=MLf+MC∅

ΔH=2kg×25kcal/kg+ 2kg×(0.6kcal/kg.K×(-14-⁻114)

=50kcal+120kcal

=170kcal

3 0
3 years ago
Pls answer 50 points for an answer!!!!
lyudmila [28]

ummmn hi i dont know lol

5 0
3 years ago
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