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Scilla [17]
3 years ago
9

A Ray of light in air makes an angle of 45 degree at the surface of a sheet of ice . the ray is refracted within the Ice at an a

ngle of 30 degree . a) what is the critical angle for the Ice? b) A speck of dirt is embedded 2CM below the surface of the Ice . what is the Apparent depth when viewed at normal incidence​
Physics
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

Explanation:

a )

Refractive index of ice = sini / sin r where i is angle of incidence and r is angle of refraction

angle of incidence = 90 - 45 = 45

angle of refraction = 30

refractive index of ice   μ  = sin 45 / sin 30

= 2/ √2

μ  = √2 = 1.414

μ  = 1 / sin C where C is critical angle for ice - air interface

sinC = 1 / μ

= 1 / 1.414

= .70

C = 44 .42° .

b )

real depth / apparent depth = refractive index

2 / apparent depth = 1.414

apparent depth = 2 / 1.414

= 1.414 cm .

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Explanation:

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3. What are the challenges of looking for Dyson spheres?
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Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud
FromTheMoon [43]

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

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wavelength = 0.6 m

Speed = frequency x wavelength

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Let the tension is T.

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Option (A) is correct.

7 0
3 years ago
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
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