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3 years ago

Light enters a substance from air at 30 degrees to thenormal.

Physics

1 answer:

Gwar [14]3 years ago

5 0

Answer:

Explanation:

The angle of incidence and refraction are both measured from the normal

angle of incidence = 30°

angle of refraction = 23°

refractive index(n) = sini / sinr

n = sin30°/sin23°

n = 1.27965

refractive index (n) = 1/sinC

where C is the critical angle.

sinC= 1/n

C =arcsin (1/n)

C =arcsin (1/1.27965)

C = 51.39°

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Explanation:

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Gre4nikov [31]

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

$\omega = \frac{v}{R}$

Where,

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R = the radius

In the case of the initial kinetic energy:

$KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2$

Where I is the moment of inertia previously defined.

$KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2$

In the case of the final kinetic energy, we have to,

$KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$ , then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)$

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4 years ago

The speed of a certain proton is 350 km/s. If the uncertainty in its momentum is 0.100%, what is the necessary uncertainty in it

Evgesh-ka [11]

Answer:

$\Delta x = 1.807 \times 10^{-10}m$

Explanation:

mass of proton, m = 1.67 x 10^-27 kg

speed of proton, v = 350 km/s = 350,000 m/s

Momentum of proton, p = mass x speed

p = 1.67 x 10^-27 x 350000 = 5.845 x 10^-22 kg m /s

uncertainty in momentum, Δp = 0.1 % of p

Δp = $\frac{0.1\times 5.845 \times 10^{-22}}{100}=5.845 \times 10^{-25}$

According to the principle

$\Delta x\times \Delta p \geq \frac{h}{2\pi }$

where, Δx be the uncertainty in position

$\Delta x\times 5.845 \times 10^{-25}= \frac{6.634 \times 10^{-34}}{2\times 3.14}$

$\Delta x = 1.807 \times 10^{-10}m$

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3 years ago