Answer:
Your answer is D It does not need to be repeatable.
the reason for this is because C is correct.
You need to be able to experiment on something multiple times so that you can gather further data to imbedded your evidence in facts.
Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
Answer:
q₃ = -4.81 nC
Explanation:
We can use the Gauss Law here:
∅ = q/∈₀
where,
∅ = Net Flux = - 216 N.m²/C
q = total charge enclosed inside sphere = ?
∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²
Therefore,
- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²
q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)
q = - 1.91 nC
So, the total charge will be sum of all three charges:
q = q₁ + q₂ + q₃
- 1.91 nC = 1.74 nC + 1.16 nC + q₃
q₃ = - 1.91 nC - 1.74 nC - 1.16 nC
<u>q₃ = -4.81 nC</u>
Start with what the paragraph is about and put it basically in your own words
If it's Kepler's law of equal areas you're talking about,
then the first of the four statements is true.
