Q=l^2 times R times t
Where Q - heat, I -current, R - resistance and t is time
If you increase I twice (and it's squared), than Q gonna went up 4 times (2 squared).
Choose last
Answer:
78g
Explanation:
Given parameters:
Mass of oxygen gas = 16g
Mass of potassium oxide = 94g
Unknown:
Mass of reacting potassium = ?
Solution:
To solve this problem, we need to obtain a balanced reaction equation. Then determine the number of moles of the reactant and use it to find that of the other one.
Balanced equation:
4K + O₂ → 2K₂O
Number of moles of reacting oxygen;
Number of moles = 
molar mass of O₂ = 2 x 16 = 32g/mole
Number of moles =
= 0.5mole
From the reaction equation;
4 mole of K reacted with 1 mole of O₂;
x mole of K will react with 0.5 mole of O₂
Therefore, 4 x 0.5 = 2 moles of potassium.
Mass of potassium = number of moles x molar mass
Molar mass of potassium = 39g
Mass of potassium = 2 x 39 = 78g
The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:

where

is the orbital radius of the first satellite

is the orbital radius of the second satellite

is the orbital period of the first satellite

is the orbital period of the second satellite
If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite: