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3 years ago

If a flea can jump straight up to a height of 0.550 m , what is its initial speed as it leaves the ground?

1 answer:

7 0

Let u = upward initial speed of the flea.

At the maximum height of h = 0.550 m, the vertical velocity is v = 0.
Therefore, from the formula
v² = u² - 2gh,
obtain
0 = u² - 2*(9.8 m/s²)*(0.55 m)
u² = 10.78 m²/s²
u = 3.283 m/s

Answer: The initial upward speed is 3.28 m/s (nearest thousandth)

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Ten students stand in a circle and are told to make a transverse wave. What best describes the motion of the students? Each stud

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Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.

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The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.

The students can make a transverse wave by raising their hands up and then down, one student at a time.

The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.

Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.

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4 years ago

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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.