This is a free fall and in free fall we use this formula:
d = (1 ÷ 2) × g × t², where d is the distance, g is the gravitational acceleration and t is the time.
In our case,
We are already given the moon's gravitational acceleration and we are going to substitute it with g. Let's leave the unknown alone, which is t.
t = √(2d ÷ g)
If we perform the formula, t is found to be √(2d ÷ g) = √(2 × 1.2 ÷ 1.62) ≅ 1.217 s
I am sorry for my bad English and if there is anything that you do not understand please let me know.
At r = 2R> R The expression for the electric field will be given by: (2R)^2*E=kQ. Where, k=(9*10^9)N.m/C^2, Q=(8*10^-10)C and R=0.025m. So substituting and clearing, we have that the magnitude of the electric field will be: E=(9*10^9)*(8*10^-10)/((2*0.025)^2)=2880 N / C.
Answer:
1.38*10^18 kg
Explanation:
According to the Newton's law of universal gravitation:

where:
G= Gravitational constant (6.674×10−11 N · (m/kg)2)
ma= mass of the astronaut
mp= mass of the planet

so:

Gravitational potential energy is caused when an object is resting above the ground. It is released when the object is falling, not by burning substances.
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]