Answer:
correct is d) a ’= g / 2
Explanation:
For this exercise let's use the kinematics equations
On earth
v = v₀ - a t
a = (v₀- v) / T
On planet X
v = v₀ - a' t’
a ’= (v₀-v) / 2T
Let's substitute the land values in plot X
a’= a / 2
Now let's use Newton's second law
W = ma
m g = m a
a = g
We substitute
a ’= g / 2
So we see that on planet X the acceleration is half the acceleration of Earth's gravity
Answer:
an air mass is a volume of air defined by its temperature and water vapor content. Air masses cover many hundreds or thousands of miles, and adapt to the characteristics of the surface below them. They are classified according to latitude and their continental or maritime source regions. Colder air masses are termed polar or arctic, while warmer air masses are deemed tropical. Continental and superior air masses are dry while maritime and monsoon air masses are moist. Weather fronts separate air masses with different density (temperature and/or moisture) characteristics. Once an air mass moves away from its source region, underlying vegetation and water bodies can quickly modify its character.When winds move air masses, they carry their weather conditions (heat or cold, dry or moist) from the source region to a new region. When the air mass reaches a new region, it might clash with another air mass that has a different temperature and humidity. This can create a severe storm.
Air masses can affect the weather because of different air masses that are different in temperature, density, and moisture. When two different air masses meet a front forms. This is one way air masses effect our weather.
To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.
By Hook's law we know that force is defined as,
Where,
k = spring constant
x = Displacement change
PART A) For the case of the spring constant we can use the above equation and clear k so that
Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m
PART B) In the case of speed we can obtain it through the period, which is given by
Re-arrange to find \omega,
Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to
Therefore the mass of the trailer is 4093.55Kg
PART C) The frequency by definition is inversely to the period therefore
Therefore the frequency of the oscillation is 0.4672 Hz
PART D) The time it takes to make the route 10 times would be 10 times the period, that is
Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s
From the average speed you can fix an equation:
Average speed = distance / time
You know the average speed = 65.1 kg / h, then
65.1 = distance / total time,
where total time is the time traveling plus 22.0 minutes
Call t the time treavelling and pass 22 minutes to hours:
65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1
From the constant speed, you can fix a second equation
Constant speed = distance / time traveling
94.5 = distance / t ==> distance = 94.5 * t
The distance is the same in both equations, then you have:
[t +22/60] * 65.1 = 94.5 t
Now you can solve for t.
65.1t + 22*65.1/60 = 94.5t
94.5t - 65.1t = 22*65.1/60
29.4t = 23.87
t = 23.87 / 29.4
t = 0.812 hours
distance = 94.5 km/h * 0.812 h = 76.7 km
Answers: 1) 0.81 hours, 2) 76.7 km
