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kozerog [31]
4 years ago
15

Two competing models attempt to explain the motions and changing brightness of the planets: Ptolemy's geocentric model and Coper

nicus' heliocentric model. Sort the characteristics according to whether they are part of the geocentric model, the heliocentric model, or both solar system models.
Drag the appropriate items to their respective bins.
Epicycles and deferents help explain planetary motion.
Planets move in circular orbits and with uniform motion.
The brightness of a planet increases when the planet is closest to Earth.
This model is Earth-centered.
This model is Sun-centered.
Retrograde motion is explained by the orbital speeds of planets.
Retrograde motion is explained by epicycles.
Epicycles and deferents help explain planetary motion. Selected
Dragable
A. Geocentric:
B. Heliocentric:
C. Both geocentric and heliocentric:
Physics
1 answer:
AlekseyPX4 years ago
6 0

Answer:

A. Geocentric: This model is Earth Centered . Retrograde motion is explained by epicycles .

B. Heliocentric: This model is Sun centered.  Retrograde motion is explained by the orbital speeds of planets

C. Both geocentric and heliocentric: Epicycles and deferents help explain planetary motion . Planets move in circular orbits and with uniform motion . The brightness of a planet increases when the planet is closest to Earth.

Explanation:

The principle of the Ptolemy's geocentric model was developed on the assumption that the center of the universe is the Earth. On the other hand, the principle of the Copernicus' heliocentric model was based on the assumption that the center of the universe is the sun. However, both models have a common ideology on uniform circular motion and epicycles.

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A current of 5.52 a flows along a wire with ℓ = (1.382 i − 2.095 j) m. the wire resides in a uniform magnetic field b = (−0.350
Masteriza [31]

Magnetic force on a wire inside magnetic field is given by formula

F = i (L X B)here we will haveℓ = (1.382 i − 2.095 j) mB = (−0.350 j + 0.550 k) t.i = 5.52 Anow as per above formula we have[tex]F = 5.52 * (1.382 i − 2.095 j) X (−0.350 j + 0.550 k)

F = 5.52* (-0.4837 k - 0.7601 j - 1.152 i)

F = - 6.36 i - 4.196 j - 2.670 k

So magnetic force on the wire is given by above expression with all three components shown with i, j, k.

x component of force is 6.36 N

y component of force is 4.196 N

z component of force is 2.67 N

8 0
3 years ago
If a 110-W lightbulb emits 2.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions. P
pogonyaev

Answer:

9.7 x 10¹¹ .

Explanation:

2.5 % of 110 W = 2.75 J/s

energy of one photon

= hc / λ

=\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}

= .036 x 10⁻¹⁷ J

No of photons emitted

= 2.75 / .036 x 10⁻¹⁷

= 76.38 x 10¹⁷

Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance

photons passing per unit area of this sphere

= 76.38 x 10¹⁷  / 4π ( 2.8)²

Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing

= \frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2

= 9.7 x 10¹¹ .

4 0
3 years ago
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s
Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

KE_i = KE_f

KE_f = \frac{1}{2} mv^2

Here,

m = Mass

V = Velocity

Replacing,

KE_f = \frac{1}{2} (12000)(11)^2

KE_f = 72600J

Therefore the  final kinetic energy of the two car system is 72.6kJ

8 0
4 years ago
What is the momentum of an 18-kg object moving at 0.5 m/s?
Otrada [13]

sorry Idk the answer ..???

6 0
3 years ago
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