Answer:
the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s
Explanation:
given information
car's mass, m = 1200 kg
= 100 m
= 
= 150 m
= 0
according to conservative energy
the distance from point A to B, h = 150 m - 100 m = 50 m
the initial speed 
final speed
= 0
thus,
² =
² - 2 g h
0 =
² - 2 g h
² = 2 g h
= √2 g h
= √2 (9.8) (50)
= 31.3 m/s
I only know #2 and #4.
2.) cells
3.) cells, life , existing
Sorry that i dont know the rest but i took a test on this not to long ago, and i tend to forget stuff once i take a test on it.
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>
- We have the expression for the initial velocity as,

- Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

- Now, the velocity will become,

<h3>How to find the speed of the satellite?</h3>
- As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Learn more about the equations of planetary motion here:
brainly.com/question/28108487
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C) alternating current .
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B)direct current </span>
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