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Elden [556K]
3 years ago
10

The speed of light in a material is 0.50

Physics
1 answer:
vivado [14]3 years ago
4 0

Answer:

Explanation:

This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum

Let's find the refractive index of the material

            n = c / v

           n = 3 10⁸ / 0.5 10⁸

           n = 6

When the material passes from one medium to another, it must comply with the law of refraction

           n₁ sin θ = n₂ sin θ₂

for the angle criticize the angles tea2 = 90

          tea = sin⁻¹n₂/ n₁

The vacuum replacement index is n₂ = 1

         tea = sin⁻¹ (1/6)

         tea = 9.59º

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Considerando que você comece a caminhar em velocidade constante, inicialmente a 350 m de um ponto referencial escolhido. Você ca
ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

S(t)=S_o+vt    ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

S(t)=350-(1m/s)t

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

             (1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

4 0
3 years ago
Compare the properties of Titan’s atmosphere with those of Earth’s atmosphere.
Aloiza [94]

Answer and Explanation:

Comparison between the Titan's atmosphere and earth atmosphere

  • Titan atmosphere is more denser than the earth atmosphere
  • The quantity of nitrogen is more in titan atmosphere than earth atmosphere titan atmosphere have about 98 % of nitrogen in the other hand earth atmosphere has only 78 %
  • There are no oxygen present in titan atmosphere while in earth atmosphere it is present
  • Its air is not suitable for breath but in earth atmosphere we can breath
5 0
3 years ago
Which is a example of application of a scientific concept?
stira [4]
I think its c or d but im not quite sure.....
3 0
3 years ago
A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands
seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0
3 years ago
As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax
AysviL [449]

Answer:

I=2.766\ kg.m^2

Explanation:

We have:

diameter of the wheel, d=0.88\ m

weight of the wheel, w_w=280\ N

mass of hanging object to the wheel, m_o=6.32\ kg

speed of the hanging mass after the descend, v_o=4\ m.s^{-1}

height of descend, h=2.5\ m

(a)

moment of inertia of wheel about its central axis:

I=\frac{1}{2} m.r^2

I=\frac{1}{2} \frac{w_w}{g}.r^2

I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

I=2.766\ kg.m^2

3 0
3 years ago
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