Answer:
A) and B) are correct.
Explanation:
If the object is at rest, it means that no net force is exerted on it.
As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.
This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).
At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.
So, the statements A) and B) are true.
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
Answer:
0.57 m
Explanation:
First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is
And the distance to cover is
d = 19 m
So the time taken is
Now we want to find how high the ball is at that time. The initial vertical velocity is
So the vertical position of the ball at time t is
where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find
The crossbar height is 3.05 m, so the difference is
So the ball passes 0.57 m above the crossbar.
Displacement is simply the change in position, or the difference in the final and initial positions:
Then
(a) ∆<em>d</em> = 5 m - 0 m = 5 m
(b) ∆<em>d</em> = 1 m - (-2 m) = 1 m + 2 m = 3 m
(c) ∆<em>d</em> = 2 m - (-2 m) = 2 m + 2 m = 4 m
(d) ∆<em>d</em> = 6 m - 2 m = 4 m
