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
Product of my research: 1 HP = 746 watts .
12 HP = (12 x 746 W) = 8,952 W
8,952 W = 8.952 kW
The correct answer is B the total velocity is equal at both landing and launch because before your about launch you have 0 velocity then when you have landed you also have 0 velocity. Hope This Helps
Part a)
At t = 0 the position of the object is given as

At t = 2

so displacement of the object is given as

so average speed is given as

Part b)
instantaneous speed is given by


now at t= 0

at t = 1


at t = 2

Part c)
Average acceleration is given as



Part d)
Now for instantaneous acceleration
As we know that

at t = 0

at t = 1

now we have

At t = 2 we have



<em>so above is the instantaneous accelerations</em>