Answer:
mass of the planet: 
Explanation:
When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):
where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:
Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.
We know the orbital radius R (
, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.
We know that the moon makes a full circumference (
) in 388800 seconds, therefore its tangential velocity is:
where we rounded the velocity to one decimal.
Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.
Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True
2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6
3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
Answer:
Explanation:
For the problem, we should have same reynolds number
ρvd/mu = constant
1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600
d = 25.66 cm
Answer:
m = 1.82E+23 kg
Explanation:
G = universal gravitational constant = 6.67E-11 N·m²/kg²
r = radius of orbit = 72,600 km = 7.26E+07 m
C = circumference of orbit = 2πr = 4.56E+08 m
P = period of orbit = 12.9 d = 1,114,560 s
v = orbital velocity of satellite Jim = C/P = 409 m/s
m = mass of Xandar = to be determined
v = √(Gm/r)
v² = [√(Gm/r)]²
v² = Gm/r
rv² = Gm
rv²/G = m
m = rv²/G
mG = universal gravitational constant = 6.67E-11 N·m²/kg²
r = radius of orbit = 72,600 km = 7.26E+07 m
C = circumference of orbit = 2πr = 4.56E+08 m
P = period of orbit = 12.9 d = 1,114,560 s
v = orbital velocity of satellite Jim = C/P = 409 m/s
m = mass of Xandar = to be determined
v = √(Gm/r)
v² = [√(Gm/r)]²
v² = Gm/r
rv² = Gm
rv²/G = m
m = rv²/G
m = 1.82E+23 kg
Correct answer: D). Working Memory
The short term memory is the part of the memory system in which the information is stored for 30 seconds. However, it can be increased by rehearsal. Short term memory is also called active memory or the working memory. The working memory is the part of the cognitive system that is responsible for storing information temporarily for processing. It is important for reasoning.
Hence, the correct answer would be option D.
