To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.
In turn, we will resort to the application of Newton's second law.
PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

Where,
X = Desplazamiento
V = Velocity
t = Time
In this case there is no initial displacement or initial velocity, therefore

Clearing for time,



PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:
F = ma
Where,
m=mass
a = acceleration
Acceleration can also be written as,

Then





Negative symbol is because the force is opposite of the direction of moton.
PART C) Acceleration through kinematics equation is defined as




The gravity is equal to 0.8, then the acceleration is


The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

The same for the diameter,

The volume of a cylinder is given as




Therefore the volume would be 
Answer:
Fluorescent lighting is usually used.
Explanation:
Answer:
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- <u><em>1,500 kg.m/s</em></u>
Explanation:
First, arrange the information in a table:
Object Mass (kg) Velocity (m/s)
A 200 15
B 150 - 10
After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.
According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.
<u>Momentum before the collision, P₁</u>:


<u>Momentum after the collision</u>:
- As stated, it es equal to the momentum before the collision: 1,500 kg . m/s
<h2>
Answer: Gravitational attraction will be the same</h2>
According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:
(1)
Where:
is the module of the force exerted between both bodies
is the universal gravitation constant.
and
are the masses of both bodies.
is the distance between both bodies
Now, if we double both masses and the distance also doubles, this means:
and
will be now
and 
will be now
Let's rewrite the equation (1) with this new values:
(2)
Solving and simplifying:
(3)
As we can see, equation (3) is the same as equation (1).
So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>