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Basile [38]
3 years ago
7

A ray in oil (n = 1.52) reaches a boundary with water (n = 1.33) at 55.9 deg. Does it reflect internally or refract into the air

?
Physics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

Ray reflects internally

Explanation:

Given:

Ray in oil (n1) = 1.52

Ray in water (n2) = 1.33

θ = 55.9°

Computation:

Step 1 . Find critical angel(θc)

Sinθc/Sin90° = Ray in water / Ray in oil

Sinθc/1 = 1.33/1.52

θc = 61°

Since, θ = 55.9° is smaller than θc = 61°

So, Ray reflects internally.

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Within a period of the periodic table, how do the properties of the elements vary?
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Answer:

Elements in the same period have the same number of electron shells; moving across a period (so progressing from group to group), elements gain electrons and protons and become less metallic. This arrangement reflects the periodic recurrence of similar properties as the atomic number increases.

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5 0
3 years ago
The mass of a coin is measured to be 12.5±0.1 g. The diameter is 2.8±0.1 cm and the thickness 2.1 ±0.1 mm. Calculate the average
Evgesh-ka [11]

The average density of the material from which the coin is made is 9.67 g/cm³.

<h3>Volume of the coin</h3>

The volume of the coin at the given diameter is calculated as follows;

V = Ah

where;

  • A is area of the coin
  • h is the thickness of the coin

V = πd²/4 x h

V = π(2.8)²/4  x (0.21 cm)

V = 1.293 cm³

<h3>average density of the coin</h3>

The average density of the material from which the coin is made is calculated as follows;

density = mass/volume

density = 12.5 g / (1.293 cm³)

density = 9.67 g/cm³

Thus, the average density of the material from which the coin is made is 9.67 g/cm³.

Learn more about average density here: brainly.com/question/1354972

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7 0
2 years ago
2. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10-9 C and the oth
Maru [420]

Answer:

A. -2.16 * 10^(-5) N

B. 9 * 10^(-7) N

Explanation:

Parameters given:

Distance between their centres, r = 0.3 m

Charge in first sphere, Q1 = 12 * 10^(-9) C

Charge in second sphere, Q2 = -18 * 10^(-9) C

A. Electrostatic force exerted on one sphere by the other is:

F = (k * Q1 * Q2) / r²

F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²

F = -2.16 * 10^(-5) N

B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:

Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))

= - 6 * 10^(-9) C

Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C

Hence the electrostatic force between them is:

F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²

F = 9 * 10^(-7) N

7 0
3 years ago
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