Answer:
because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits
Answer:
Vagetagble, fruits,fresh milk and juice vitamins c
To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,


Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity



PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,


From the speed it is possible to use find the formula, so



Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.
PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is



Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
The figure shown above is series combination as the two resistors (bulb) are there which are then connected to the battery
so i conclude from the above options given the option is B
hope it helps