Answer:
the interaction between the use of a charged object
Explanation:
honestly just had my certified physics educator dad tell me to write that lol sorry.
Answer:
182.5 s
Explanation:
From the law of conservation of angular momentum,
I₁ω₁ = I₂ω₂
where I₁,ω₁ are the rotational inertia and angular speed of the star and I₂,ω₂ are the rotational inertia and angular speed of the white dwarf star
I₁ = 2/5MR₁² where M = mass of star and R₁ = radius of star = radius of sun = 696340 km
I₂ = 2/5MR₂² where M = mass of white dwarf star = mass of star and R₂ = radius of white dwarf star = radius of earth = 6400 km
ω₁ = 2π/T₁ where T₁ = period of star = 25 days = 25 × 24 × 60 × 60 s = 2.16 × 10⁶ s
ω₂ = 2π/T₂ where T₂ = period of white dwarf star.
So, I₁ω₁ = I₂ω₂
2/5MR₁² × 2π/T₁ = 2/5MR₂² × 2π/T₂
R₁²/T₁ = R₂²/T₂
T₂ = T₁R₂²/R₁² = 2.16 × 10⁶ s × (6400 km/696340 km)² = 182.5 s
I think least electricity is used between probably 7-8a.m. and 4-5p.m.
This is because, around those times, the suns already out. Depending on how sunny it is, it may not be as cold as all the other times of the day. And by then, buildings are typically already warmed all up. Everybody's body heat also may play a factor in buildings. ( if there is a ton of people )
Weight equals mass multiplied by the acceleration due to gravity.