AFA – 2022 – EN
AFA – 2022
Compare these and other solutions from AFA-2022 and from the entrance exams of other universities on fisicaevestibular.com.br with others, and you will see that:
It has better visuals and more enlightening illustrations.
It is intended for students who truly struggle with understanding Physics and Mathematics, always striving to explain even the smallest details.
It does not just list formulas but always aims to show their origins and uses.
It seeks to explain through drawings and illustrations.
It does not skip steps but explains each sequence step by step.
There is a great concern that someone understands the solutions. The teacher puts themselves in the student’s place.
It often provides information different from what is necessary for the solutions, but you will find useful information in the upcoming portals.
And much, much more.
Air Force Academy (AFA ) is a higher education institution of the Brazilian Air Force , located in Pirassununga , State of São Paulo and recognized by the Ministry of Education .
It is part of the training and improvement system for the Command ‘s personnel .
the Air Force (COMAER) and is subordinate to the Education Department (DEPENS) of the Brazilian Air Force (FAB).
Its purpose is to train active officers for the FAB’s aviators, quartermasters and infantry cadres.
It is considered one of the three best military pilot training schools in the world and trains not only military pilots for the Brazilian Air Force, but also for Air Forces in countries in Latin America, Africa, Asia and Europe, through international cooperation agreements.
Admission to the AFA is subject to approval in annual national public admissions exams. Information about these exams is available at all organizations of the Air Force Command.
Currently, the Academy trains cadets in the following courses:
Air Force Infantry Officers Training Course (CFOINF)
Aviation Officer Training Course (CFOAV)
Quartermaster Officer Training Course (CFOINT).
Furthermore, all Academy cadets, at the end of the course, receive a bachelor’s degree in Administration, with an emphasis on Public Administration.
01 – (AFA-2022)
In Figure 1, below , there is a top view of a uniform circular motion described by two particles, A and B, which travel along semicircular paths, with radii , respectively,
on a table, always keeping them aligned with center C.
When they reach the edge of the table, as illustrated in Figure 2, the particles are thrown horizontally and describe parabolic trajectories, free from any resistance forces, until they reach the floor, which is flat and horizontal.
During this fall, particles A and B travel horizontal distances, respectively .
Angular velocity (W) of an MCU
Relationship between scalar (V) and angular (W) velocity
Horizontal launch
By placing the origin of the reference system at the launch point, for example, the X axis is oriented to the right and the Y axis downwards.
At each instant, the movement is broken down into two parts:
According to the Y axis, it is a uniformly varied movement with zero initial vertical velocity , that is, it is a free fall with the body abandoned from the origin, subject only to the acceleration of gravity, of intensity g, vertical direction and downwards. 
Equations:
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02 – (AFA-2022)
The following graphs representing rectilinear movements were presented to a physics student .
When analyzing the graphs, the student realized that they can represent the same movement, the graphs
a) I and II, only.
b) I and III only.
c) II and III, only.
d) I , II and III.
Graph (I) refers to the time-lapse graph of space (position) of a uniform motion (MU) which is a first-degree function (inclined line) with zero acceleration.
Graph (II) refers to the hourly velocity graph (first-degree function, inclined line) with
constant acceleration of uniformly varied motion (MUV).
Graph (III) also refers to a MUV where the acceleration is constant (line parallel to the time axis).
Therefore graphs (II) and (III) refer to a uniformly varied movement (MUV)
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03 – (AFA-2022)
A candidate for the Air Force Officer Training Course, after passing all previous stages, must take a Physical Fitness Assessment Test (TACF).
One of the TACF tests consists of running 2,000m within a maximum time interval.
To do so, the candidate will complete 5 laps on a track consisting of two straight sections of length L and two semicircular sections of radius R, always staying on the dotted line, as illustrated in the following figure.
On his first lap, the candidate travels the semicircular sections at constant speed V and
Under these conditions, the straight section L of this track has a length, in m, equal to
a) 50
b) 250
c) 100
d) 400
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04 – (AFA-2022)
Two blocks, A and B, of negligible dimensions are dropped, starting from rest, from the top of a plane inclined at 30 ° to the horizontal; traveling, after the same interval of time, the indicated distances as illustrated in the following figure.
If you don’t master the theory, see the summary below:
Inclined plane with friction
Data are:
A- A
05 – (AFA-2022)
A homogeneous beam 3 m long is in equilibrium, attached to the wall through points A and B, as illustrated in the following figure .
At point A, there is a frictionless joint that allows the beam to rotate freely.
At point B, an ideal spring 1, whose deformation is x, connects the beam to the wall.
A load P hangs, through an ideal thread, at the end C of the beam and is at a height of 2 m in relation to the free end of an ideal spring 2, vertically fixed to the horizontal floor, as can also be seen in the figure.
At a given moment, the wire is cut and P falls, without experiencing air resistance, onto the trimmer, of negligible mass, causing spring 2 to undergo a deformation of 40 cm until it stops.
Knowing that sen θ = 0.6, cos θ = 0.8 and that the elastic constants of spring 1 and 2 are equal, it can be stated that the deformation x, of spring 1, in cm, before the wire was cut, was equal to
a) 7.5
b) 25
c) 40
d) 50
During the fall of the block, since friction is neglected you can apply the mechanical energy conservation theorem whose theory is summarized below:
Mechanical Energy
Conservative systems
After the wire is cut, the block with weight P falls from point R, without experiencing air resistance, onto the trimmer, of negligible mass, causing spring 2 to undergo a deformation of x = 40 cm = 04 m until it stops at point S.
Thus, placing the height reference level at point S, the block falls moving from R to S from a height h = 2.0 + 0.4 = 2.4 m.
P force weight of the block transmitted to point C through the string
If the system is in equilibrium, you must use the moment (torque) theorem of each force in rotational equilibrium: “The algebraic sum of the moments (torques) of the forces acting on the system in relation to a point chosen pole O, around which the beam tends to rotate, must be zero”.
To calculate the moment (torque) of each force, carefully observe the figure below, where we consider the clockwise direction of rotation around the pole as positive and the counterclockwise direction as negative.
R- ?
06 – (AFA-2022)
For the tension in the wire to be zero , the ratio between the volume of the bar that is submerged only in the liquid with density and its total volume can be expressed as
In solving this exercise you must use Archimedes’ Buoyancy Theorem, whose theoretical summary is below:
Statement of Archimedes’ principle
Mathematical expression of Buoyancy
Mathematical expression of weight (vertical and downward force)
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07 – (AFA-2022)
Relative humidity provides the degree of water vapor concentration in an environment.
When this concentration reaches 100% (which corresponds to saturated vapor), condensation occurs.
Relative humidity (RH) is obtained by comparing the density of water vapor in the air with the density of the vapor if it were saturated, i.e.
The following table gives the maximum water vapor concentration measured at the indicated temperatures.
On a short day with a temperature of 32 °C and a relative humidity of 40%, a person notices that a glass of cold soda begins to condense water vapor (becomes “sweaty”).
Under these conditions, the temperature, in °C, of the glass of soda was, at most ,
a) 5
b) 10
c) 15
d) 20
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08 – (AFA-2022)
Considering that the heat loss from this environment occurs only through the door, the power, in W, of a heater capable of maintaining this temperature constant must be equal to
Data: K = 0.8 W/mK
a) 1200
b) 2400
c) 3200
d) 4800
If you haven’t mastered the theory, here it is:
Fourier’s law
The above relationships are expressed by Fourier’s law through the equation
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09 – (AFA-2022)
To inflate the tire of his bicycle, a cyclist, as shown in the figure below, has a cylindrical pump, whose cross-sectional area (a) is equal to 20 cm².
The connecting hose (M) is non-deformable and has negligible volume.
The bicycle’s front tire has a volume of 2.4 L and initially has an internal pressure of 0.3 atm.
The internal pressure of the pump, when the piston (E) is fully pulled to a height (H) of 36 cm, is equal to 1 atm (normal atmospheric pressure).
Consider that, during calibration, the tire volume remains constant and that the process is isothermal, with an ambient temperature of 27°C.
Under these conditions, to raise the tire pressure to 6.3 atm, the number of repetitions that the cyclist must do, moving the piston to the end of its stroke, is
a) 20
b) 50
c) 80
d) 95
Clapeyron equation
Cylinder
Tire
A- A
10 – (AFA-2022)
A projectile with mass 2m is fired horizontally with a speed of module v, as shown in Figure 1, and moves at this speed until it collides with a simple pendulum of length L and mass m, initially at rest, in a perfectly elastic collision.
Consider that the projectile was launched from a distance very close to the pendulum and that, after the collision, this pendulum begins to oscillate in simple harmonic motion, as indicated in Figure 2, with amplitude A.
Neglecting the action of dissipative forces, the oscillation period of this pendulum, immediately after the collision, is given by
Calculation of the velocity of the sphere of mass m after the collision using the conservation of momentum and the coefficient of restitution. See theoretical summary below:
Principle of Conservation of Momentum
In the case of exercise:
Before the collision the velocity of the sphere of mass 2m is V and that of the sphere of mass m is zero.
Energy at MHS
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11 – (AFA-2022)
An optical arrangement, represented by Figure 1, consists of a two-dimensional luminous object aligned with the optical and geometric center of a support S that can be occupied individually.
Consider that all graphic elements that can be installed on the support are ideal and that the arrangement is immersed in air.
A- A
12 – (AFA-2022)
A source emits two types of electrically charged particles, which are launched into a region where only a uniform vertical electric field E acts . 
These particles penetrate perpendicularly to the field, from point A, with speed , colliding with a vertical barrier at points S and R, as illustrated in the figure .
Characteristics of the Electric Field Vector
Intensity (module) of the resulting force on each charge:
A- A