The formula for the acceleration due to gravity is:
a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet
So, if they have the same a:
m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
m₂ = 4m
<em>Thus, the answer is D.</em>
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

Recall that the speed of light is equivalent to

Then replacing,


Our values are given as




Replacing we have,



Therefore the magnetic field around this circular area is 
Answer:
lead with Z = 82 is transformed into Bismuth with Z = 73
The Po with Z = 84 becomes Pb with Z = 82
Explanation:
Beta decay occurs when a neutron emits an electron and an anti neutrino from the atomic nucleus, therefore the atomic number of the material increases by one unit.
Pb + e + ν → Bi
lead with Z = 82 is transformed into Bismuth with Z = 73
In alpha decay, a helium nucleus is emitted from the nucleus of the atom, therefore the atomic number decreases by two units.
Po +α → Pb
The Po with Z = 84 becomes Pb with Z = 82
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

Answer:
Savage is the answer
Explanation: The savage people guess like me