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Vaselesa [24]
3 years ago
6

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

action of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its n

Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

\beta = 41.68°

Explanation:

according to snell's law

\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }

refractive index of water n_w is 1.33

refractive index of glass  n_g  is 1.5

sin\alpha = \frac{n_w}{n_g}* sin30

sin\alpha = 0.443

now applying snell's law between air and glass, so we have

\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}

sin\beta = \frac{n_g}{n_a} sin\alpha

\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]

we know that sin\alpha = 0.443

\beta = 41.68°

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
3 years ago
A mid-ocean ridge under the ocean is MOST similar to a _____ on land.
Alexxx [7]
A mid ocean ridge is a under water mountain range, knowing this what would you say it is most similar to on land? hope this helped!
5 0
2 years ago
The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas
Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 m,/sec^2

(c) Escape velocity on earth is 3.563\times 10^4m/sec

Explanation:

We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

Mass of earth M_E=5.972\times 10^{24}kg

So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

Volume of mars V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3

So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

(A) So the ratio of mean density \frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735

(B) Value of g on mars

g is given by g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2

(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

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I have no clue. Sorry bro
6 0
2 years ago
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The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.
kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

P = 300\ N \ \times  \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\&#10;P = 300 \ N \times 22.22 \ m/s\\\\&#10;P = 6,666 \ W

The power of the engine at the given efficiency is calculated as follows;

E = \frac{P_{out}}{P _{in}} \times 100\%\\\\&#10;60\% = \frac{P_{out}}{6,666} \times 100\%\\\\&#10;0.6 = \frac{P_{out}}{6,666} \\\\&#10;P_{out} = 3,999.6 \ W

Learn more about efficiency here: brainly.com/question/15418098

8 0
2 years ago
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