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

are their multiple choices ?

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards