From the gravity acceleration theorem due to a celestial body or planet, we have that the Force is given as

Where,
F = Strength
G = Universal acceleration constant
M = Mass of the planet
m = body mass
r = Distance between centers of gravity
The acceleration by gravity would be given under the relationship


Here the acceleration is independent of the mass of the body m. This is because the force itself depended on the mass of the object.
On the other hand, the acceleration of Newton's second law states that

Where the acceleration is inversely proportional to the mass but the Force does not depend explicitly on the mass of the object (Like the other case) and therefore the term of the mass must not necessarily be canceled but instead, considered.
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
Answer:
Electric field due to two charges is given as

Explanation:
As we know that two charges are opposite in nature
So the electric field at the mid point of two charges will add together
so the net field is given as

now we have


now we have

