PLEASE HELP!!!!

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure $(P_1)$ =210+101.325=311.325

$T_1=25^{\circ}\approx 298 K$

$V=0.025 m^3$

$T_2=50^{\circ}\approx 323 K$

as the volume remains constant therefore

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{311.325}{298}=\frac{P_2}{323}$

$P_2=337.44 KPa$

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass $m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}$

$m_1=0.91 kg$

Final mass $m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}$

$m_2=0.839$

Therefore $m_1-m_2$ =0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0

3 years ago

Engineers tasked with building a car bumper need high-quality plastic that is readily available. They found the material polycar

jenyasd209 [6]

Answer:

The material must be durable (quality of the material requirement)

Explanation:

The design criteria set for the materials used for technological design are;

1) The materials should be affordable (less costly)

2) The materials should be last for a long duration (high durability)

3) The material should be readily available (easily sourced)

Therefore, given that the engineers initially had the criteria for the required plastic to be of high quality and to be readily available, and that the poly-carbonate they found is long lasting and not too costly, the criteria met that was set initially was the quality criteria of durability.

7 0

3 years ago

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I need help with part D

natka813 [3]

1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.

8 0

3 years ago

What happens to the water particles in a wave?

lys-0071 [83]

Answer: waves transport energy, not water. As a wave crest passes, the water particles move in circular paths. The movement of the floating inner tube is simulacra to the movement of the water particles. Water particles rise as a wave crest approaches.

Explanation:

8 0

3 years ago

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