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DedPeter [7]
3 years ago
13

What are the layers of the earth

Physics
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

CRUST, UPPER MANTLE, LOWER MANTLE, INNER CORE, OUTER CORE,

Explanation:

how do u not know this

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Estás cuentas son ingresos fijos o variables o egresos fijos o variables?
Brut [27]
¿El salario es un costo fijo o variable?
Los salarios anuales son costos fijos, pero otros tipos de compensación, como comisiones o horas extraordinarias, son costos variables.
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3 years ago
Corresponde al conjunto de puntos por donde pasa un cuerpo al moverse
Anna35 [415]

Answer:

Means?

Explanation:

6 0
3 years ago
A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a
andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

sin\theta = 1.22\frac{\lambda}{d}

Here,

\lambda = Wavelength

d= Diameter of aperture

\theta = Angular resolution or diffraction angle

Our values are given as,

\theta = 11\°

The frequency of the sound is f = 9100 Hz

The speed of the sound is v = 343 m/s

The wavelength of the sound is

\lambda = \frac{v}{f}

Here,

v = Velocity of the wave

f = Frequency

Replacing,

\lambda = \frac{(343 m/s)}{(9100 Hz)}

\lambda = 0.0377 m

The diffraction condition is then,

sin\theta = 1.22\frac{\lambda}{d}

Replacing,

sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}

d = 0.24 m

Therefore the diameter should be 0.24m

6 0
3 years ago
What’s a real life example of relative motion?
vesna_86 [32]

Answer:

let's say you're on a bus going 50 km/hr, you are moving at a velocity of zero relative to the bus. however, relative to the ground you are moving at the same velocity as the bus.

Explanation:

physics

7 0
3 years ago
Explain diffraction at a single slit (light)
Leto [7]

Answer:

At some point on say, the receiving screen, light emanating from the left side of the slit will be out of phase (a difference of 1/2 wavelengths) from   light coming from the center of the slit.

Thus for every point that is left of the center of the slit, there will be a point on the right side of the slit that is out of phase,

There will be no light on the screen at that particular point and thus there will be a dark fringe there.

That is the basic explanation for the appearance of dark and bright fringes on the receiving screen.

4 0
2 years ago
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