Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
<span>The answer should be the vegitation. </span>
PART A)
horizontal distance that will be moved = 14 m
Height of the fence = 5.0 m
height from which it is thrown = 1.60 m
angle of projection = 54 degree
So here we can say that stone will travel vertically up by distance

now we will have displacement in horizontal direction

now we know that


now we will have


also for y direction


now from the two equations we will have




now from above equations


So the minimum speed will be 13.2 m/s
Part B)
Total time of the motion after which it will land on the ground will be "t"
so its vertical displacement will be

now we will have




Now the time after which it will reach the fence will be t1 = 1.8 s
so total time after which it will fall on other side of fence


now the displacement on the other side is given as



Answer:
2.08 kg
Explanation:
Newton's second law states that the acceleration of an object is proportional to the force applied to the object, according to the equation:

where F is the force applied, m is the mass of the object and a its acceleration.
In this situation, the soccer ball is kicked with a force F=13.5 N and its acceleration is a=6.5 m/s^2, therefore its mass is

I'm assuming you want the first law of thermodynamics.
The First Law of Thermodynamics states that heat is a form of energy and cannot be created or destroyed. It can, however, be transferred from one location to another and can be converted into other forms of energy.