The electrostatic force between two charges is inversely
proportional to the square of the distance between them.
So if you want to multiply the force by, say, ' Q ',
you need to multiply the distance by ( 1 / √Q ) .
We want to multiply the force by 16, so we need to
multiply the distance by ( 1 / √16 ) = ( 1 / 4 ) .
The distance should be changed to 1/4 of what it is now.
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
A closed system is a system that is completely isolated from its environment. The physical universe, as we currently understand it, appears to be a closed system. An open system is a system that has flows of information, energy, and/or matter between the system and its environment, and which adapts to the exchange.
D makes the most sense, but you just have to put two and two together and go with your gut feeling, first cross out the answers that don't make sense (A didnt make sense) and go from there! I hope the little bit above me helped you answer or decide :) Good luck!
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
