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timurjin [86]
3 years ago
6

When monochromatic light passes through the interface between two unknown materials at an angle θ where 0∘<θ<90∘, no chang

es in the direction of propagation of light are observed. What can be said about the two materials?
Physics
1 answer:
aliina [53]3 years ago
4 0

The question is incomplete, the options in the question are:

The two materials have matching indexes of refraction.

The second material through which light propagates has a lower index of refraction.

The second material through which light propagates has a higher index of refraction.

The two materials are identical

Answer:

The two materials have matching indexes of refraction.

Explanation:

The phenomenon of refraction reveals that, when light propagates through the interface between two materials with different indexes of refraction, the direction of propagation of light changes. When this does not happen, it means that the two materials have very similar refractive indexes. For instance, the refractive index of glass Pyrex is 1.474 and that of glycerin is 1.473 hence no relevant change in the direction of propagation of light can be observed when light passes between the interface of the two substances. That does not mean that they are both the same material or identical materials.

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Lead-202 has a half-life of 53,000 years. How long will it take for 15/16 of a sample of lead-202 to decay?
ohaa [14]

Answer:

C. 212,000 years

Explanation:

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4 0
3 years ago
A 0.0414 kg ingot of metal is heated to 243◦C
tino4ka555 [31]

Answer:

448 J/kg/°C

Explanation:

m₁ C₁ (T₁ − T) + m₂ C₂ (T₂ − T) = 0

(0.0414 kg) C (243°C − 20.4°C) + (0.411 kg) (4186 J/kg/°C) (18°C − 20.4°C) = 0

(9.22 kg°C) C − 4129 J = 0

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3 0
3 years ago
If the load on Pulley E is 20N, how much effort is needed to life it?
Mariana [72]
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7 0
3 years ago
The figure shows the arrangement of master cylinder and slave cylinder of a part of braking system. The master cylinder piston,
Neko [114]

Answer:

a) 35 kPa

d) 140 N

c) 1) Increasing the brake fluid  pressure

2)  Increasing the slave piston surface area.

Explanation:

The parameters given are;

a) Force applied to the master cylinder piston = 28 N

Cross sectional area of the master cylinder piston = 8 cm² = 8 × 10⁻⁴ m²

The pressure P on the brake fluid is given by the formula for pressure as follows;

P= \dfrac{Applied \ force}{Area \over \ which \  force \ is \ applied} = \dfrac{28 \, N}{8 \times 10^{-4} \, m^2} = 35,000 \, N/m^2 = 35,000 \, Pa

The pressure on the brake fluid, P, produced by the master cylinder piston = 35,000 Pa = 35 kPa

b) Given that the area of the slave piston = 40 cm² = 0.004 m², we have from the formula for pressure, P;

P= \dfrac{Applied \ force}{Area \over \ which \  force \ is \ applied} = \dfrac{Applied \ force}{4 \times 10^{-3} \, m^2} = 35,000 \, N/m^2

Therefore;

Applied force on the slave piston = 4 × 10⁻³ m² × 35,000 N/m² = 140 N

c) The force, F produced by the slave cylinder piston is given by the relation;

F = Pressure × Area

Therefore, the two ways of increasing the force produced by the slave cylinder piston is as follows;

1) Increasing the pressure in the brake fluid by increasing the force exerted by the master cylinder piston

2) Increasing the surface area of the slave piston.

4 0
3 years ago
Step 3 - Fillin the missing numbers:<br> 21 -3 -10 24<br> b d
mash [69]

Answer:

b = 2

c = -1

d = -2

Explanation:

Got this question right on edge

4 0
3 years ago
Read 2 more answers
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