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neonofarm [45]
3 years ago
12

For 589-nm light, calculate the critical angle for the following materials surrounded by air. (a) fluorite (n = 1.434) ° (b) cro

wn glass (n = 1.52) ° (c) ice (n = 1.309)
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

a) θc = 44.21⁰.

b) θc = 41.14⁰

c)  θc = 40.36⁰

Explanation:

sin θ = \dfrac{n_2}{n_1}

For air,

n₂ = 1

so,

sin θ =  \dfrac{1}{n_1}

(a) fluorite : n₁ = 1.434⁰

sin θ = 0.697                   θc = 44.21⁰.

(b) crown glass: n₁ = 1.52⁰  

sin θ = 0.657                   θc = 41.14⁰.

(c) sodium chloride : n₁= 1.544⁰

sin θ = 0.647                   θc = 40.36⁰.

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Gravity affects weight of an object

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Communication with submerged submarines via radio waves is difficult because seawater is conductive and absorbselectromagnetic wa
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Answer:

D. 4000 km

Explanation:

f = Frequency of wave that is being transmitted = 76 Hz

\lambda = Wavelength of wave that is being transmitted

v = The velocity of electromagnetic waves through air = 3\times 10^8\ m/s

Velocity of a wave is given by

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8 0
3 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

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