where would the spaceprobe experience the strongest net (or total) gravitional force exerted on it by Earth and Mars

r = 41.1 10⁹ m

Explanation:

For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal  

                  ∑ F = 0

                  F (Earth- probe) - F (Mars- probe) = 0

                  F (Earth- probe) = F (Mars- probe)

Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system

the distance from Earth to the probe is        R (Earth-probe) = r

the distance from Mars to the probe is        R (Mars -probe) = D - r

where D is the distance between Earth and Mars

                 M_earth (D-r)² = M_Mars r²

                 (D-r) =    r

                  r ( ) = D

                  r =

We look for the values ​​in tables

                  D = 54.6 10⁹ m (minimum)

                  M_earth = 5.98 10²⁴ kg

                  M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg

let's calculate

                  r = 54.6 10⁹ / (1 + √(0.642/5.98)  )

                  r = 41.1 10⁹ m

answer

c, on a nylon carpet in a dry area. give me brainliest

explanation:

what type of charge does it have? negative