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3 years ago

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3. How did the light wave interact with the water? | I

1 answer:

fredd [130]3 years ago

4 0

Answer:

Refraction is another way that waves interact with matter. ... Waves bend as they enter a new medium because they start traveling at a different speed in the new medium. For example, light travels more slowly in water than in air. This causes it to refract when it passes from air to water or from water to air.

Explanation:

i hope this helps but if it doesnt im rlly srry :p

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Is renewable energy just as efficient as fossil fuels?

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Answer:

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Explanation:

First of all it's more than efficient than fossil fuels

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3 years ago

A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.

REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}}$

$= \frac{g \sin \theta}{1 +\frac{2}{3}}$

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= 0.6 x 9.81 x sin ( 29.7)

$= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

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Δt = 1.411 s

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3 years ago

Pube Goldberg machine is a complicated contraption designed to do a very simple task, like the one shown above

Elanso [62]

Answer:

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Explanation:

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3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

Umnica [9.8K]

Answer:

<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

$\displaystyle w=\sqrt{\frac{k}{m}}$

The equation for the motion of the object is written as a sinusoid:

$\displaystyle X=A\ cos\ w\ t$

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

$\displaystyle X'=V=-A\ w\ sin\ w\ t$

And the maximum speed is

$\displaystyle V_{max}= A\ w$

Solving for the amplitude

$\displaystyle A= \frac{V_{max}}{w}$

Computing w

$\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s$

Calculating A

$\displaystyle A=\frac{1.7}{24.5}=0.069\ m$

$\displaystyle \boxed{A=6.9\ cm}$

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4 years ago