Answer:
To find the position at which a 68.0-kg object would experience a net force of zero due to two other objects, we need to consider the gravitational force acting on the object due to each of the other two objects and the distance between them.
Let's assume that the two other objects have masses of M1 and M2, and are located a distance of d1 and d2 away from the 68.0-kg object, respectively. The force of gravity between the 68.0-kg object and each of the other two objects can be calculated using Newton's Law of Gravitation:
F1 = G * M1 * m / d1^2
F2 = G * M2 * m / d2^2
where G is the gravitational constant, m is the mass of the 68.0-kg object, and F1 and F2 are the gravitational forces acting on the 68.0-kg object due to the other two objects.
To find the position where the net force on the 68.0-kg object is zero, we need to find the values of d1 and d2 that satisfy the equation:
F1 = F2
Substituting the equations for F1 and F2, we get:
G * M1 * m / d1^2 = G * M2 * m / d2^2
Simplifying the equation:
d2^2 / d1^2 = M1 / M2
Taking the square root of both sides:
d2 / d1 = sqrt(M1 / M2)
Now, we can choose any value for d1, and then solve for d2 using the above equation. For example, let's say we choose d1 to be 1 meter. Then:
d2 = d1 * sqrt(M1 / M2) = 1 * sqrt(M1 / M2)
So, the position where the net force on the 68.0-kg object is zero is at a distance of 1 meter from one of the other two objects, and at a distance of sqrt(M1 / M2) meters from the other object.
Step-by-step explanation:
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