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Volgvan
3 years ago
7

The critical angle for refraction from glass to air is 30° and that from water to air is 37°.Find the critical angle for refract

ion from glass to water.​
Physics
1 answer:
erik [133]3 years ago
8 0

Answer:

Å

1,55

5 400 Å

n

=

193548,4 km/s

1,55

3 10 m/s

n

c

Vidrio v: =

3 724,14 Å

1,45

5 400 Å

n

=

206 896,5 km/s

1,45

3 10 m/s

n

c

Aceite = v:

07,1

45,1

55,1

n

n

I

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In the desert enviroment the chemical weathering of rocks is generally reduced because
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8 0
3 years ago
A force F acts in the forward direction on a cart of mass m. A friction force ff opposes this motion. Part A Use Newton's second
stellarik [79]

Answer:

a = F-ff/m

Explanation:

According to Newton's second law of motion which states that "the rate of change in momentum of a body is directly proportional to the applied force F and acts in the direction of the force.

Mathematically;

F = ma

Since two forces acts on the cart i.e the moving force F and the frictional force Ff , we will take the sum of the forces.

∑F = ma where

m is the mass of the cart

a is its acceleration

∑F = F+(-ff )(since frictional force is an opposing force)

F - ff = ma

Dividing both sides by mass m

a = F-ff/m

5 0
3 years ago
If a liquid has a density of 1.67 g/cm^3, what is the volume of 45g of the liquid?
Aloiza [94]

Answer:

V = 26.95 cm³

Explanation:

Density is given by the formula :

ρ = m÷V

Density = mass ÷ Volume

Given both density and mass we rearrange, substitute and solve for Volume :

Rearranging the equation to make Volume the subject :

ρ = m÷V

ρV = m

V = m÷ ρ

Now substitute :

V = 45 ÷ 1.67

V = 26.9461077844

Take 2 decimal places as the density is 2 decimal places :

V = 26.95

Units will be cm³ as it is volume

Hope this helped and have a good day

8 0
2 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
3 years ago
Which of the following statements is true?
viva [34]
Statement B is correct.
4 0
3 years ago
Read 2 more answers
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