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3 years ago

Select the correct answer.

Physics

1 answer:

Vesna [10]3 years ago

3 0

Answer:

A. continental-oceanic convergent

Explanation:

I knew it couldn't be B because it's oceanic and continental, not oceanic and oceanic.

Next, I noticed the word convergent, which implies "coming together" to me.

I looked it up and noticed the term convergent referred to a plate boundary where a plate slips under (subducted) another, so I knew it was A.

Hopefully, this helps you understand the question better. Have a great day!

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A solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with anangular speed of 3.5 rad/s. Wha

Lynna [10]

Answer:

kE=0.0735 J

Explanation:

Given that

Radius ,R=10 cm = 0.1 m

Mass ,m= 3 kg

Angular speed ,ω = 3.5 rad/s

We know that moment of inertia for solid sphere given as

$I=\dfrac{2}{5}mR^2$

Kinetic energy

$KE=\dfrac{1}{2}I\omega^2$

Now by putting the values

$KE=\dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \omega^2$

$KE=\dfrac{1}{2}\times \dfrac{2}{5}\times 3\times 0.1^2\times 3.5^2\ J$

kE=0.0735 J

Therefore the kinetic energy will be 0.0735 J

3 0

3 years ago

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.