<h2>
Answer:g=9.79
,A object of mass

at the surface of earth experiences a force

</h2>
Explanation:
Let
be the mass of earth.
Let
be the radius of earth.
Let
be the universal gravitational constant.
Given,




Let
be the acceleration due to gravity.
Then,


A object of mass
at the surface of earth experiences a force 
Answer:
Some examples of vegetative propagation are farmers creating repeated crops of apples, corn, mangoes or avocados through asexual plant reproduction rather than planting seeds. Vegetative propagation can be accomplished from side-shoots, slips, stems and sections of tubers, bulbs or rhizomes.
Explanation:
Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m
Answer:
0.661 s, 5.29 m
Explanation:
In the y direction:
Δy = 2.14 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.661 s
In the x direction:
v₀ = 8 m/s
a = 0 m/s²
t = 0.661 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²
Δx = 5.29 m
Round as needed.