The variables which are involved in understanding Kepler's third law of
motion are
<h3 /><h3>What is Kepler's third law of motion?</h3>
Kepler's third law of motion states that the the square of the orbital period of
a planet is proportional to the cube of the semi-major axis of its orbit. He
also inferred that the greater the distance, the slower the orbital velocity.
This thereby makes option D the most appropriate option as it contains the
orbital velocity and distance to sun variables.
Read more about Kepler's third law of motion here brainly.com/question/777046
The correct answer to your question here is D
Answer: The green house effect is best described by option 4 (Energy given off by earth is reflected off of earth's atmosphere back down to the surface).
Explanation:
The green house effect can be described as the energy given off by earth is reflected off of earth's atmosphere back down to the surface.
When energy from the sun passes through the atmosphere, some are absorbed which keeps the earth surface warm. While the rest is reflected back largely by cloud.
The energy which is emitted from the earth surface is called the infrared radiation. Some of the infrared radiation passess through the atmosphere but most is absorbed and re- emitted in all directions by the greenhouse gas molecules and clouds. This effect warms the earth surface and the lower atmosphere. Therefore this statement (Energy given off by earth is reflected off of earth's atmosphere back down to the surface) is correct about greenhouse effect.
For the greenhouse effect to occur, greenhouse gas molecules are mostly needed. Examples of these gases include:
--> Carbon dioxide (CO2),
--> Water vapor (H2O), and
--> Methane (CH4)
Over the years, the excessive human activities has lead to increase in the greenhouse gas molecules which has negatively affected the greenhouse effects.
Answer:
option B is the correct answer
Explanation:
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Answer:
3.6 seconds
Explanation:
Given:
y₀ = y = 0 m
v₀ = 31 sin 35° m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²
0 = 17.78t − 4.9t²
0 = t (17.78 − 4.9t)
t = 0 or 3.63
Rounded to the nearest tenth, the ball lands after 3.6 seconds.