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Sati [7]
3 years ago
9

True or False: Amplitude can increase or decrease wavelength

Physics
1 answer:
Marat540 [252]3 years ago
4 0

Answer:

A. False, frequency can increase or decrease wavelength.

For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.

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The side length of a square is modeled by s=A, where A is the area of the square. Graph the function. What is the side length of
Naya [18.7K]

Answer:

√35 square units

Explanation:

Area of square = 35square units

Formula of area of square is a =s*s where a is area and s is side

So,

s*s = 35

s^2= 35

s = √35

tell me if this is the answer u r looking for

3 0
2 years ago
The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how hi
sweet-ann [11.9K]
<span>373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km</span>
7 0
3 years ago
Read 2 more answers
A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the
olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

f = frequency of the tone = 329.6 Hz

T = Time period of the sound wave

we know that, Time period and frequency are related as

T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s

v = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m

v = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m

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7 . . . . . the amplitude
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