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>
Increase the change in momentum of the collision
Answer:
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Explanation:
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Answer:
Keeping the speed fixed and decreasing the radius by a factor of 4
Explanation:
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"
It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,
R' = R/4
New centripetal acceleration will be,




So, the centripetal acceleration of the ball can be increased by a factor of 4.