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Troyanec [42]
2 years ago
7

Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh

t travels faster in X. (ii) The light will bend towards the normal. (iii) The light will speed up. (iv) The light will bend away from the normal. a. (ii) b. (iii) and (iv) c. (i) d. (i) and (ii) e. (ii), (iii) and (iv)
Physics
1 answer:
ahrayia [7]2 years ago
8 0

Answer:

i) FALSE,  ii) TRUE,  iii) FALSE, iv)  FALSE

Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

           n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

Let's apply these equations to our case where

          nₓ <n_y

i) The expression of the refractive index

          nₓ < n_y

         \frac{c}{v_x} = \frac{c}{n_y}

         v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

 the statement of FALSE

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A 0.20 kg mass is oscillating at a small angle from a light string with a period of 0.78 s.
Scorpion4ik [409]

Answer:

L = 15 cm

Explanation:

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(0.78/2π)²

L = 0.151027... m

L = 15 cm

7 0
2 years ago
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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli
Masja [62]

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
La mascota de felipe esta jugando en el parque despues de un tiempo esta agotado de tanto correr. Felipe conoce que el perro con
MatroZZZ [7]

Responder:

6.704 m / s

Explicación:

Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.

Trabajo realizado = Fuerza × distancia

Distancia = Trabajo realizado / Fuerza

Distancia = W / mg

S = 176/8 × 9,81

S = 176 / 78,48

S = 2,24 m

Dada la velocidad inicial u = 3.6km / h

Convertir a m / s

= 3.6km × 1000m / 1h × 3600

= 3600/3600

= 1 m / s

u = 1 m / s

Usando la ecuación de movimiento

v² = u² + 2gS para obtener la velocidad final v:

v² = 1² + 2 (9,81) (2,24)

v² = 1 + 43,9488

v² = 44,9488

v = √44,9488

v = 6,704 m / s

Por tanto, la rapidez final del perro es de 6,704 m / s

6 0
2 years ago
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