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pickupchik [31]

2 years ago

15

When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is

travelling from media 1 to media 2.
At what angle of incidence is the reflected ray perpendicular to the incident ray? The indexes of refraction for the two media are n1 and n2, respectively.

Express your answer in terms of the variables n1 and n2

1 answer:

podryga [215]2 years ago

3 0

Answer: critical angle, sin^-1 (n2/n1)

Explanation: the angle of incidence at which the retracted ray makes an angle of 90° with the normal is known as the critical angle.

Snell's law defined refraction mathematically as shown below

n1 sin θi = n2 sin θr

n1 = refractive index of the first medium

n2 = refractive index of the second medium

θi = angle of incidence

θr = angle of refraction

When the refrafted ray is perpendicular to the normal, the angle of refraction (θr) is 90° hence making the angle of incidence (θi) the critical angle θc

By substituting these conditions into the Snell's law, we have that

n1 sin θc = n2 sin 90

According to trigonometry, the value of sin 90 is 1, hence we have that

n1 sin θc =n2

sin θc = n2/n1

θc = sin^-1 (n2/n1)

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A neutral atom of tin (Z = 50) has 69 neutrons. (a) How many protons does it have? (b) What is its mass number?

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Answer:

Protons = 50

Mass number = 119

Explanation:

A neutral of tin, Sn, has a Z value of 50 and neutrons are 69.

Z is the atomic number. The atomic number is equal to number of protons or number of electrons in an atom that is neutral i.e:

Z = PROTONS = ELECTRONS

So the number of protons = 50 since Z is 50

>>>>>>>>>>>>>>>¦

The bulk of the mass of an atom is concentrated in the nucleus. To calculate mass number, we simply sum all the elementary particles in the nucleus i.e protons and neutrons. These particles are called the nucleons.

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Is it possible for a gas contained in a chamber to maintain a constant temperature while heat is being added to the gas? Explain

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Answer:Yes

Explanation:

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Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

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2 years ago

Read 2 more answers