F = ma
F = (1000 kg)•(5 m/s^2)
F = 5000 N
Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock: 
Pebble: 
(b)Calculate the magnitude of the acceleration of each object when released.
Rock: 
Pebble: 
Explanation:
The universal law of gravitation is defined as:
(1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
<em>Case for the rock </em>
<em>:</em>
m1 will be equal to the mass of the Earth
and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth
.

Newton's second law can be used to know the acceleration.

(2)

<em>Case for the pebble </em>
<em>:</em>


Answer:
k = 44000 N/m
Explanation:
Given the following data;
Maximum gravitational potential energy = 770 J
Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J
Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m
To find the spring constant, k;
The elastic potential energy of an object is given by the formula;
Substituting into the equation, we have;
Cross-multiplying, we have;
k = 44000 N/m
Refer to the diagram shown below.
Define

= unit vector in the eastern direction

= unit vector n the northern drection.
Then the displacement vectors are
Vector #1:

Vector #2:

Vector #3:

Because the vector sum of all three vector is zero, therefore
1381.1 +0.6561a - 0.891b = 0
703.7 - 0.9873a + 0.454b = 0
That is,
0.6561a - 0.891b = -1381.1 (1)
-0.9873a + 0.454b = -703.7 (2)
From (1), obtain
b = 0.7364a + 1550.1 (3)
Substitute (3) into (2).
-0.9873a + 0.454(0.7364a + 1550.1) = - 703.7
-0.653a = -1407.4
a = 2155.3 mles
From (3), obtain
b = 3137.2 mles
Answer:
The magnitudes of the two displacement vectors are
2155.3 miles and 3137.2 miles.