Johannes Kepler is known for which discovery in astronomy? discovering planets around other stars
determining planetary laws of motion
recording 20 years of data about planetary motion
proposing the heliocentric model of the solar system
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Answer:
Determining planetary laws of motion
Explanation
Kepler worked for Tycho Brahe who was recording planetary motion for 20 years. After Brahe's demise Kepler inherited those records of planetary motion. After analyzing those records he put forth the three laws of planetary motion.
Law of Orbits: Every planet in our solar system move in elliptical orbit with sun at one focus It is also called the law of Ellipses
Law of Areas: The imaginary line drawn between the planet and sun will sweep out equal areas in equal period
Law of periods: The square of the time period of the planet is directly proportional to the cube of the semi major axis of the orbit. It is also called the law of Harmonies
Johannes Kepler is known for determining the planetary laws of motion in astronomy. Therefore option 2 is correct.
"determining planetary laws of motion," is the correct answer. Johannes Kepler was a German astronomer and mathematician who made significant contributions to our understanding of the motion of planets.
His three laws of planetary motion, known as Kepler's laws, revolutionized the field of astronomy and laid the foundation for Isaac Newton's later work on universal gravitation.
Kepler's first law, known as the law of ellipses, states that planets orbit the Sun in elliptical paths, with the Sun at one of the foci of the ellipse. This challenged the prevailing notion that planetary orbits were perfect circles.
Know more about Kepler's first law:
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1. Before dropping, all potential energy
2.dropping, potential energy transformed to kinetic energy
3. before hitting the ground, all Kinetic energy.
Recall the formula for both energy, which are U=mgh, and K=1/2mv^2
Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2
plug the number in and solve for velocity.
2x400x10=1/2 x 2 x v^2
v^2=8000
v=
v=40
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The direction of the pull is -70.93°, or approximately 71° South of West.
Given the vector as F= (−2980.0i^ + 8200.0j^)N
The magnitude of the pull is given by the formula below:
Magnitude of pull, |F| = √(F_x^2 + F_y^2)
Here, F_x is the force in the x-direction and F_y is the force in the y-direction. We know that:
F_x = -2980.0 N and F_y = 8200.0 N
Therefore, the magnitude of the pull is:
|F| = √(F_x^2 + F_y^2)|F| = √((-2980.0)^2 + (8200.0)^2)|F| = √(88840000) |F| = 9425.89 N
The direction of the pull can be found as follows:
θ = tan⁻¹(F_y/F_x)θ = tan⁻¹(8200.0/-2980.0)θ = -70.93°
The direction of the pull is -70.93°, or approximately 71° South of West.
learn more about pull on:
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b. 75%
c. 50%
d. 0% E. It cannot be determined.
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you can assume that the relative humidity is 100% . . . since
no water was able to evaporate from the wet bulb.