Answer:
The ability of our bodies to adapt to different levels of gravity. You would become weaker and your heart is use to zero gravity. Boredom because there isn't much to in space. When intelligent people get bored, it's not pretty all the time...
Answer:
The frictional force
6.446 N
The acceleration of the block a = 6.04 
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by


3.9 × 9.81 × 
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by

F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by

23.554 = 3.9 × a
a = 6.04 
This is the acceleration of the block.
Answer: it’s A and B
Explanation: everyone else on this post was giving you the wrong answer.
Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is
<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N
which is positive, so it's directed east.
To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that
18 N + <em>F</em>₃ = -12 N
<em>F</em>₃ = -30 N
So <em>F</em>₃ has a magnitude of 30 N and points west.
Answer: 
Explanation:
Given
Radius of flywheel is 
Angular acceleration 
For no change in radius, tangential acceleration is given as

Insert the values
