The gravitational force of two objects, by definition, is given by:
Where,
G: gravitational constant
m1: mass of object number 1.
m2: mass of object number 2.
d: distance between both objects.
Therefore, according to the given equation, a change that always results in an increase in gravitational force is:
Increase in the mass of the objects and decrease in the distance between them.
Answer:
A change that will always result in an increase in the gravitational force between two objects is:
Increase in the mass of the objects and decrease in the distance between them.
Here's a useful factoid that you don't hear about very often:
1 volt means 1 Joule per Coulomb.
When 1 coulomb of charge falls or gets lifted through 1 volt potential difference, it gains or loses 1 Joule of energy.
If you want to lift 5 coulombs to a height of 1 volt, you have to give it 5 joules.
If you actually give those 5 coulombs 7.5 joules instead, they'll rise up to 1.5 volts above the potential where they started. The flowed through a potential DIFFERENCE of 1.5 volts.
(If they started at a point that's connected to the Earth, like a water pipe or a metal flagpole, then their new potential is 1.5 volts, because we define zero as the potential of the ground.)
Answer:
Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. There are many types of waves including sound waves, water waves and all the parts of the electromagnetic spectrum.
Explanation:
The fusion reaction that is easiest to accomplish is the reaction between two hydrogen isotopes: deuterium, extracted from water and tritium, produced during the fusion reaction through contact with lithium. When deuterium and tritium nuclei fuse, they form a helium nucleus, a neutron and a lot of energy.
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
