Answer:
At the point of dropping the object, by Newton's first law due to gravitational force = m × g, accelerates
By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)
As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out
The body then moves at a constant uniform motion down according to Newton's first law
Explanation:
At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g
As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force
Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.
Substract two consecutive terms of the sequence to see if there is a common difference:
As we can see, there is a common difference of -6.
Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).
Notice that the first term of the sequence is 3.
Then, the rule for the sequence is to start with 3 and add -6 repeatedly.
Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.
Answer:
Approximately .
Explanation:
Consider this slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin
.
Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as
.
Convert the initial speed of this diver to SI units:
.
The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at (
near the surface of the earth.) At
seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:
To eliminate from this expression, solve the equation between
and
for
. That is: express
as a function of
.
.
Replace the in the equation of
with this expression:
.
Plot the two functions:
and look for their intersection. Refer to the diagram attached.
Alternatively, equate the two expressions of (right-hand side of the equation, the part where
is expressed as a function of
.)
,
.
The value of can be found by evaluating either equation at this particular
-value:
.
.
The position vector of a point on a cartesian plane is
. The coordinates of this skier is approximately
. The position vector of this skier will be
. Keep in mind that both numbers in this vectors are in meters.
