Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer:
460 g
Explanation:
Heat lost by the warm water = heat gained by the cold water
-mCΔT = mCΔT
-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)
-m (37°C − 85°C) = (1000 g) (37°C − 15°C)
-m (-48°C) = (1000 g) (22°C)
m = 458 g
Rounded to two significant figures, you need a mass of 460 g of water.
This group is called “noble gases” because they do not react with other elements. This is because they have a full valence shell.
Due to the fact that no one can consume .04 of a tablet, we can round down this answer to 1. This means that Mr. Jones should take C- 1 tablet per day.
I hope I've helped! :)
