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Maslowich
3 years ago
5

Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy

stem. have the Sun at their exact center. are highly inclined to the ecliptic. are almost circular, with low eccentricities.
Physics
1 answer:
BaLLatris [955]3 years ago
4 0

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

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A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi
S_A_V [24]

Answer:

10.2 Watt

Explanation:

N  = number of turns in flat coil = 160

A  = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

B  = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

t  = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

E = \frac{- N A \Delta B}{t}

E = \frac{- (160) (0.20) (- 0.80)}{2}

E = 12.8 volts

R = Resistance of the coil = 16 Ω

Power is given as

P = \frac{E^{2}}{R}

P = \frac{(12.8)^{2}}{16}

P = 10.2 Watt

6 0
3 years ago
An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish
Snezhnost [94]

Answer:

7.1 m/s

Explanation:

First, find the time it takes for the fish to reach the water.

Given in the y direction:

Δy = 6.1 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.12 s

Next, find the velocity needed to travel 7.9 m in that time.

Given in the x direction:

Δx = 7.9 m

a = 0 m/s²

t = 1.12 s

Find: v₀

Δx = v₀ t + ½ at²

7.9 m = v₀ (1.12 s) + ½ (0 m/s²) (1.12 s)²

v₀ = 7.1 m/s

4 0
3 years ago
Read 2 more answers
A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each
nataly862011 [7]

The centripetal acceleration is 13.0 m/s^2

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference (2\pi r) and the period of revolution (T), so it can be rewritten as

v=\frac{2\pi r}{T}

Therefore we can rewrite the acceleration as

a=\frac{4\pi^2 r}{T^2}

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

T=\frac{1}{4}s = 0.25 s

Substituting into the equation, we find the acceleration:

a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0
3 years ago
Please follow my sis on i.n.s.t.a.g.r.a.m Id - tehzz_13
abruzzese [7]

Answer:

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4 0
3 years ago
Read 2 more answers
An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to
bulgar [2K]

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

P = 2\pi r

Where,

r= radius

While centripetal acceleration is defined by

a=\frac{v^2}{r}

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

\frac{P}{4} = \frac{2\pi r}{4}

The same route you take is the distance traveled, that is

x = \frac{P}{4}

x = \frac{2\pi r}{4}

1725 = \frac{2\pi r}{4}

r = 1098.17m

PART B)

With the radius is possible calculate he centripetal acceleration,

a = \frac{v^2}{r}

a = \frac{115^2}{1098.17}

a = 12.04m/s^2

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of 12.04m/s^2

3 0
3 years ago
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