Answer:
F' = (4/9)F
Explanation:
The electrostatic force between two charged objects is given by Coulomb's Law:
F = kq₁q₂/r² -------------------- equation (1)
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between charges
Now, when the charges and distance altered as follows:
q₁' = 2q₁
q₂' = 2q₂
r' = 3r
Then,
F' = kq₁'q₂'/r'²
F' = k(2q₁)(2q₂)/(3r)²
F' = (4/9)kq₁q₂/r²
using equation (1):
<u>F' = (4/9)F</u>
Answer:
if i were you i would try to do the work because if you let someone else you wont be able to understand the question
The answer is c so the area can grow
The process of splitting one large nucleus into
smaller ones is nuclear fission.
The process of combining two small nuclei into
one larger one is nuclear fusion.
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
