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oksano4ka [1.4K]
3 years ago
12

Once the lander has successfully landed on the surface, what is the minimum amount of thrust required to set it into motion agai

n? Based on your answer, what can you conclude about the moon’s gravity?
Physics
1 answer:
arsen [322]3 years ago
7 0

Answer:

The exact performance you need is dependent on a large number of engineering factors, but around 2:1 initial thrust to weight

Explanation:

The acceleration due to gravity on the surface of the Moon is approximately 1.625 m/s2, about 16.6% that on Earth's surface or 0.166 ɡ. Thus the intitial interia doesn't change from when on the moon compared to earth but rather in free-fall when inside a space capsule.

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What is dark energy and what evidence do astronomers have that it is an important component of the universe?
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Answer:

The existance of dark energy was suggested to help explain measurements, using Type Ia supernovae as distance indicators, that the expansion of the universe is speeding up and its acceleration requires a source of energy. Scientists suggested that It may be a new form of energy for which there is not yet a theoretical explanation. quantum mechanics predict that it may be the vacuum energy associated with "empty" space itself.

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4 years ago
It is possible a tree could fall down, even if no one did anything to it. A. True B. False
hichkok12 [17]

Answer:

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The tree can fall down even though no one did anything to it...

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3 years ago
Who actually asked Abraham to sacrifice his son?
pochemuha

I think god did ??? I searched it up okay

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3 years ago
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You want to move a 4- kg bookcase to a different place in the living room. If u push with a force of 65 n and the bookcase accel
IrinaK [193]

Answer:

1.65

Explanation:

The equation of the forces along the horizontal direction is:

F-F_f = ma (1)

where

F = 65 N is the force applied with the push

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\mu is the coefficient of kinetic friction

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since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find

F-\mu mg = ma

And solving for \mu,

\mu = \frac{F-ma}{mg}=\frac{65-(4)(0.12)}{4(9.8)}=1.65

7 0
3 years ago
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