All categories

2 years ago

the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67

Physics

1 answer:

baherus [9]2 years ago

4 0

Answer:

$\theta_c = 36.78^o$

Explanation:

The relationship between the refractive index and the critical angle is given as follows:

$\eta = \frac{1}{Sin\ \theta_c} \\\\Sin\ \theta_c = \frac{1}{\eta}\\\\\theta_c = Sin^{-1}(\frac{1}{\eta} )$

where,

η = refractive index = 1.67

θc = critical angle =?

Therefore,

$\theta_c = Sin^{-1}(\frac{1}{1.67} )$

$\theta_c = 36.78^o$

You might be interested in

A 32-n force acts on a small mass in the positive x-direction. a 26-n force also acts on it in the negative x-direction. what is

motikmotik

Equilibrium force is the force that will keep the small mass in place, hence no movement must be made. So we know that 32 N of force is acted towards the positive direction so +32 N. Which is counteracted by 26 N force so:

32 N – 26 N = 6 N (positive)

Since positive 6 is left, therefore this must be acted by an equilibrant negative 6 N.

Answer:

6 0

2 years ago

A man pushed the box with a force of 20 N. He doubled his force to 40 N. The box did not

Alex787 [66]

Answer:

Explanation:

if the man doubles his force to 40 and the box was yet to move that means acceleration also doubled so your answer would be A

8 0

2 years ago

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

5 0

2 years ago

A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in

lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

A = $1.5 \times 2.30$

= 3.45 $in^{2}$

The given data is as follows.

Allowable stress = 14,500 psi

Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

$P_{max} = \sigma_{a}A$

= $14500 \times 3.45$

= 50025 lb

Now, maximum load from shear stress is as follows.

$P_{max} = 2 \times \tau_{a} \times A$

= $2 \times 7100 \times 3.45$

= 48990 lb

Hence, $P_{max}$ will be calculated as follows.

$P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

= 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.

4 0

3 years ago

A gun is fired parallel to the ground. at the same instant a bullet of equal size and mass next to the muzzle is released and dr

jolli1 [7]

Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>

<u></u>

<h3>What is vertical acceleration?</h3>

A vertical acceleration is typically one for which the direction of the vector is vertically upward, usually aligned with and opposite to the gravity vector. But this is a descriptive term, not a rigorous or technical term. A car may accelerate along a road and that would generally be assumed to be a horizontal.

The vector perpendicular to this direction, as perhaps a suspension motion over a bump, would be described as vertical even if it is not strictly vertical.

Note that acceleration is defined as the rate of change of the velocity vector. But the gravitation vector, ‘g’, generally vertically downward, is often denoted by what acceleration a mass in free fall (absent air resistance) would experience, i.e. the relationship between mass and weight.

Learn more about vertical acceleration

brainly.com/question/19528199

#SPJ4

3 0

10 months ago

the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67​

the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67