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hammer [34]
2 years ago
14

Calculate the critical angle between glass (n = 1.90) and ice (n = 1.31)? 43° 52° 60° 75°

Physics
1 answer:
Ket [755]2 years ago
4 0

Answer: 43 \°

Explanation:

According to <u>Snell’s Law</u>:  

n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})

Where:  

n_{1}=1.90 is the first medium index of refraction  (glass)

n_{2}=1.31 is the second medium index of refraction   (ice)

\theta_{1} is the angle of the incident ray  

\theta_{2} is the angle of the refracted ray

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

Now, the critical angle \theta_{c} is the angle from which there is no refraction and all the incident light is reflected to the same medium from which it proceeds, that is, the total internal reflection occurs. This is only possible when the index of refraction of the medium where the light strikes is higher than the index of refraction of the other medium, then the second angle (the exit angle) will reach the 90\º, for this critical incident angle \theta_{c}.

Since n_{1}>n_{1}, \theta_{1}=[tex]\theta_{c}[/tex] and \theta_{2}=90\º, hence:

n_{1}sin(\theta_{c})=n_{2}sin(90\º)

(1.90)sin(\theta_{c})=(1.31)sin(90\º)

Isolating \theta_{c}:

\theta_{c}=sin^{-1} (\frac{1.31}{1.90})

Finally:

\theta_{c}=43.5\º

The option that is close to this value is 43\º

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In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de
snow_tiger [21]

As we know that range of the projectile motion is given by

R = \frac{v^2 sin(2\theta)}{g}

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

\theta, 90 - \theta

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0
2 years ago
De Vico Comet orbits the Sun every 74.0 years and has an orbital eccentricity of 0.96. Find the comet's average distance from th
Debora [2.8K]

Answer: The comet's average distance from the sun is 17.6AU

Explanation:

From Kepler's 3rd Law, P^2=a^3

Where P is period in years

and a is length of semi-major axis or the average distance of the comet to the sun.

Given the orbital period to be 74 years

74^2 =a^3

5476 = a^3

Cube root of 5476 =a

17.626 = a

Approximately a= 17.6 AU

5 0
2 years ago
A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x
sattari [20]
If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .

I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect. On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on. The main question we're left with after all of this is: Why sardines ? ?
6 0
2 years ago
José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


8 0
2 years ago
The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis
lord [1]

Answer:

The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

Explanation:

As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.

<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>

Where,

<em>R = Orbital radius (in this case) = 30.1 AU</em>

<em />

Plug the value of R in the equation (A):

<em>(A) => The circumference of the circle = 2*π*(30.1)</em>

<em> The circumference of the circle = </em><em>60.2π</em>

Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

5 0
2 years ago
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