Answer:
3234.2 W
Explanation:
Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.
So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².
Now, the power radiated on the patch of area 0.570 m² at the equator is
P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W
Answer:
a. Planets move on elliptical orbits with the Sun at one focus.
Explanation:
Johannes Kepler was an astronomer who discovered that planets had elliptical orbits in the early 1600s (between 1609 and 1619).
The three (3) laws published by Kepler include;
I. The first law of planetary motion by Kepler states that, all the planets move in elliptical orbits around the Sun at a focus.
II. According to Kepler's second law of planetary motion, the speed of a planet is greatest when it is closest to the Sun.
Thus, the nearer (closer) a planet is to the Sun, the stronger would be the gravitational pull of the sun on the planet and consequently, the faster is the speed of the planet in terms motion.
III. The square of any planetary body's orbital period (P) is directly proportional to the cube of its orbit's semi-major axis.
Hence, one of Kepler's laws of planetary motion states that planets move on elliptical orbits with the Sun at one focus. This is his first law of planetary motion.
Answer:
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Explanation:
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Answer:
correct is d) a ’= g / 2
Explanation:
For this exercise let's use the kinematics equations
On earth
v = v₀ - a t
a = (v₀- v) / T
On planet X
v = v₀ - a' t’
a ’= (v₀-v) / 2T
Let's substitute the land values in plot X
a’= a / 2
Now let's use Newton's second law
W = ma
m g = m a
a = g
We substitute
a ’= g / 2
So we see that on planet X the acceleration is half the acceleration of Earth's gravity
Answer:
Yes, you are pulling on Earth. Reasoning. Third Newton's law of motion, action and reaction law, sates that for every action force, there is an equal (in magnitude) and opposite reaction force.
Explanation:
goo gle.