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beks73 [17]
2 years ago
14

In the picture shown below A represents a characteristic of only geocentric model, B represents a characteristic common to both

geocentric and heliocentric models, C represents a characteristic of only heliocentric model, and D represents a characteristic which the geocentric and heliocentric models do not have.
Under which label will the characteristic, "The sun and planets revolve around a central moon in the solar system" fall?
A
B
C
D
Physics
1 answer:
worty [1.4K]2 years ago
4 0
I believe the correct answer is C
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Two loudspeakers are placed on a wall 2 m apart. A listener stands directly in front of one of the speakers, 81.7 m from the wal
OverLord2011 [107]

Answer:

The phase difference is       \Delta \phi = 1.9995 rad  

Explanation:

From the question we are told that

    The distance between the  loudspeakers is d = 2m

     The distance of the listener from the wall  D = 81.7 \ m

     The frequency of the  loudspeakers is  f = 4450Hz

      The velocity of sound is v_s = 343 m/s

     

The path difference of the sound wave that is getting to the listener is mathematically represented as

        \Delta z  =\sqrt{d^2 + D^2} -D

Substituting values

        \Delta z  =\sqrt{2^2 + 81.7^2 } -81.7

       \Delta z  =0.0245m

The phase difference is mathematically represented as

           \Delta \phi =  \frac{2 \pi}{\lambda } *  \Delta z

Where \lambda is the wavelength which is mathematically represented as

          \lambda  = \frac{v_s }{f}

substituting value  

          \lambda  = \frac{343 }{4450}

        \lambda  = 0.0770 m

Substituting value into the  equation for phase difference

      \Delta \phi =  \frac{2 * 3.142 * 0.0245}{0.0770}

      \Delta \phi = 1.9995 rad  

8 0
3 years ago
A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average
Maksim231197 [3]

Answer: 72.66 AU=1.089(10)^{10} km

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit”:

T^{2}\propto a^{3} (1)  

Now, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:  

T^{2}=a^{3} (2)  

So, knowing T=619.36 years and isolating a from (2) we have:  

a=\sqrt[3]{T^{2}} (3)  

a=\sqrt[3]{(619.36 years)^{2}} (4)  

Finally:

a=72.66 AU T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km

4 0
3 years ago
How long does it take for a truck accelerating at 1.5 m/s^2 to got from rest to 75 km/hr
Free_Kalibri [48]

Answer:

t = 13.9s

Explanation:

u = 0 m/s

v = 75 km/h

= 20.83 m/s

a = 1.5 m/s²

Using

v = u + at

20.83 = 0 + 1.5t

t = 13.9s

6 0
3 years ago
Write the factors of potential energy
IgorLugansk [536]
It depends on two factors height and mass

6 0
2 years ago
Calculer l intensité du courant qui le traverse
MissTica

Answer:

Le calcul du courant se fait avec deux éléments : la tension et la valeur de la résistance. Courant (A) = tension (V) / résistance (Ohm) ce qui donne la formule I = U/R.

please mark me as brainalist

6 0
2 years ago
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