<h2>
Option 2 is the correct answer.</h2>
Explanation:
Elastic collision means kinetic energy and momentum are conserved.
Let the mass of object be m and M.
Initial velocity object 1 be u₁, object 2 be u₂
Final velocity object 1 be v₁, object 2 be v₂
Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s
Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M
Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J
Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M
We have
Initial momentum = Final momentum
24 = 3v₁ + 6M
v₁ + 2M = 8
v₁ = 8 - 2M
Initial kinetic energy = Final kinetic energy
96 = 1.5 v₁² + 18 M
v₁² + 12 M = 64
Substituting v₁ = 8 - 2M
(8 - 2M)² + 12 M = 64
64 - 32M + 4M² + 12 M = 64
4M² = 20 M
M = 5 kg
Option 2 is the correct answer.
The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
Answer:
The electronic configuration of Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5. Fe2+ contains 2 fewer electrons compared to the electronic configuration of Fe.
