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Lesechka [4]
3 years ago
13

If a sound wave is produced with a wavelength of 1.04m what is the waves frequency

Physics
1 answer:
dmitriy555 [2]3 years ago
5 0
You should just ask the wave 
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Please help me with this question :
aalyn [17]

Answer:

  • 514.27 ( wavelength )

the color is green

  • 602.93 nm  ( orange color )

the observation is that there is a change of visible color

Explanation:

A) wavelength of visible light that is most strongly reflected from a point on a soap

refraction n = 1.33

wall thickness (t) = 290 nm

2nt = (2m +1 ) ∝/2 -----equation 1

note when m = 0

therefore ∝ =  4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this

when m = 1

equation 1 becomes

∝ = 4nt/3 =( 4 * 1.33 * 290) /  3 = 1542.8 / 3 = 514.27 ( wavelength )

the color is green

B) the wavelength when the wall thickness is 340 nm

∝ = 4nt / 2m +1

where m = 1

∝ = (4 * 1.33 * 340 ) / 3  = 1808.8 / 3 = 602.93 nm  ( orange color )

the observation is that there is a change of visible color

7 0
3 years ago
During which moon phase do spring tides occur?
Murrr4er [49]

Answer:

The Full Moon and New Moon

Explanation:

6 0
3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

3 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a
timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack (v), measured in meters per second, is equal to the product of the angular speed of the wheel (\omega), measured in radians per second, and the distance of the tack respect to the rotation axis (R), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

\omega = 2.83\,\frac{rev}{s} \times \frac{2\pi\,rad}{1\,rev}

\omega \approx 17.781\,\frac{rad}{s}

Then, the tangential speed of the tack is: (\omega \approx 17.781\,\frac{rad}{s}, R = 0.393\,m)

v = \left(17.781\,\frac{rad}{s} \right)\cdot (0.393\,m)

v = 6.988\,\frac{m}{s}

The tangential speed of the tack is 6.988 meters per second.

7 0
3 years ago
1. During which phase of the moon may a solar eclipse occur? (Points : 1)
kotykmax [81]
1. New moon
This is because the moon comes between the Earth and sun and this is only possible during its new moon phase.

2. Full moon
This is because the Earth comes between the Moon and the sun and the effect is only visible when there is a full moon.

3. Corona
The corona is the outer layer and is the only one visible when there is an eclipse.
7 0
3 years ago
Read 2 more answers
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