What happen to the salt and the sand when water is added and the mixtures is stirred?

What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

Goshia [24]

Answer:

a) $R_s = 0.092$

b) $R_p = 0.085$

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

$R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2$

using snell's law to calculate θ t

$sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}$

$cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}$

a) $R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2$

$R_s = 0.092$

b) $R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2$

$R_p = 0.085$

3 0

3 years ago

A volleyball player’s hand applies a 39 N force while in contact with a volleyball for 2 seconds. What is the impulse on the bal

nikklg [1K]

Given:
F = 39 N, the force applied
t = 2 s, the time interval in which the force is applied.

By definition, the impulse is
$I = \int_{0}^{2} F(t) dt = \int _{0}^{2} 39dt = 39*2=78 \, N-s$

Answer: 78 N-s

3 0

3 years ago

Read 2 more answers

To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you

Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

β = 10 log $\frac{I}{I_o}$