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bekas [8.4K]
3 years ago
15

What is the speed of a wave that has a frequency of of 3.7 * 10 ^ 3 Hz and a wavelength of 1.2*10^ -2 m?

Physics
2 answers:
ale4655 [162]3 years ago
6 0

Answer:

44.4m/s

Explanation:

v = (3.7*10 ^3) ( 1.2 * 10 ^ -2)

= 44.4 m/s

vredina [299]3 years ago
4 0

Answer:

44.4 m/s

Explanation:

By using the wave speed formula, v = fλ ( speed= frequency × wavelength)

v = 3.7×10³ × 1.2×10^-2

= 44.4 m/s

Hope that's the answer you're looking for:)

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An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to
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Explanation:

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= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

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3 years ago
The speed of light inside a medium is (2.0 x 10^8m/s) What is the index of refraction (n) of the medium?
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Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium

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3 0
2 years ago
A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi
Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L)  ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L

Finally, using the Parallel Axis Theorem, we calculate I_B:

I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4}  ML^{2} =\frac{1}{3} ML^{2}

5 0
2 years ago
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