In a free body diagram for an object projected upwards;
- the acceleration due to gravity on the object is always directed downwards.
- the velocity of the object is always in the direction of the object's motion.
An object projected upwards is subjected to influence of acceleration due to gravity.
As the object accelerates upwards, its velocity decreases until the object reaches maximum height where its velocity becomes zero and as the object descends its velocity increases, which eventually becomes maximum before the object hits the ground.
To construct a free body diagram for this motion, we consider the following;
- the acceleration due to gravity on the object is always directed downwards
- the velocity of the object is always in the direction of the object's motion.
<u>For instance:</u>
upward motion for velocity ↑ downward motion for velocity ↓
↑ ↓
↑ ↓
acceleration due to gravity ↓
↓
↓
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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
Answer:
r = √(k q₁ q₂ / F)
Explanation:
F = k q₁ q₂ / r²
Multiply both sides by r²:
F r² = k q₁ q₂
Divide both sides by F:
r² = k q₁ q₂ / F
Take the square root of both sides:
r = √(k q₁ q₂ / F)
