Classify each of the following chemical reactions. Upper S plus upper O Subscript 2 right arrow upper S upper O subscript 2. Upp
er C a upper C l subscript 2 plus 2 upper A g upper N upper O subscript 3 right arrow upper C a (upper n upper O subscript 3) subscript 2 plus 2 upper A g upper C l. Upper Z n plus upper C u upper S upper O subscript 4 right arrow upper C u plus upper Z n upper S upper O subscript 4. 2 upper N a subscript 2 upper O right arrow 4 upper N a plus upper O subscript 2.
1 answer:
You might be interested in
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ =
q₂ =
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀