I only doing this for points, but i think its c
Answer: 0.313 rad/s
Explanation:
The equation that relates the velocity 
and the angular velocity 
in the uniform circular motion is:
(1)
Where 
is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.
Isolating from (1):
(2)
On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration 
is 
, and is given by the following equation:
(3)
Isolating :
(4)
(5)
Substitutinng (5) in (2):
(6)
This is the angular velocity that would produce an “artificial gravity” of 9 .
In the world or do you have a list?
After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.
<h3>What is Half-Life?</h3>
Half-Life refers to the time it takes for half the amount of a substance to disappear or change.
The nucleus of the atoms of radioactive elements disintegrate to half their starting amounts after every Half-Life.
After three half-lives one-eight of the original atoms remain.
Therefore, after three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.
Learn more about Half-Life at: brainly.com/question/26689704
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Answer:
see explanation
Explanation:
Given that,
velocity of 1.50 km/s = 1.50 × 10³m/s
acceleration of 2.00 ✕ 1012 m/s2
electric field has a magnitude of strength of 18.0 N/C
![\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]](https://tex.z-dn.net/?f=%5Cbar%20F%3D%20q%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%5Cbar%20B%5D%5C%5C%5C%5C%5Cbar%20F%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D%5C%5C%5C%5C%5C%5Cm%20%5Cbar%20a%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D)
![9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]](https://tex.z-dn.net/?f=9.1%20%5Ctimes%2010%5E-%5E3%5E1%20%5Ctimes%202%5Ctimes%2010%5E1%5E2%20%5Chat%20k%3D-1.6%5Ctimes10%5E-%5E1%5E9%20%5Chat%20k%20%5B18%5Chat%20k%2B%201.5%5Ctimes%2010%5E3%20%5Chat%20i%20%5Ctimes%20%28B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20k%29%5D)
