A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational force exerted on each by the earth.(b)Calculate the magnitude of the acceleration of each object when released.

A) Gravitational Force of Rock = 48.995N

Gravitational Force of Pebble = 2.94 x 10^(-3) N

B) Magnitude of acceleration of rock = 9.8 m/s²

Magnitude of acceleration of pebble = 9.8 m/s²

Explanation:

A) First of all, let's solve for the rock;

From Newton's law of universal gravitation, gravitational force is;

Fg = (Gm1m2)/r²

Where G is a constant = 6.67 × 10^(-11) Nm² /kg2²

Mass of earth (m1) = 5.97 × 10^24 kg

Radius of earth (r) = 6.38 × 10^6 m²

Mass of rock (m2) = 5kg.

Thus, Fg = [6.67 × 10^(-11) x 5.97 × 10^24 x 5] / (6.38 × 10^6)² = 48.995N

Doing the same for the pebble of mass 3 x 10^(-4) kg;

Thus, Fg = [6.67 × 10^(-11) x 5.97 × 10^(24) x 3 x 10^(-4)] / (6.38 × 10^6)² = 2.94 x 10^(-3) N

B) Let's calculate for the acceleration of the rock;

We know that Fg = ma

Thus, Let's make a the subject of the formula,

So, Fg/m = a

Substituting rock value of Fg and m to get;

48.995/5 = 9.8 m/s²

Now let's do the same for the pebble;

Fg/m = a

Substituting pebble value of Fg and m to get;

[2.94 x 10^(-3) N] / (3 x 10^(-4)) = 9.8 m/s²

according to jean piaget, in what stage of cognitive development are children able to understand conservation task?

b. preoperational

just took it