Answer:
Time is 14.8 s and cannot landing
Explanation:
This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing are the same
Calculate the time to stop, where it has zero speed
Vf² = Vo² + a t
t = - Vo² / a
t = - 110²/(-7.42)
t = 14.8 s
This is the time it takes to stop the jet
Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero
Vf² = Vo² + 2 to X
X = -Vo² / 2 a
X = -110² / 2 (-7.42)
X = 815.4 m
Since this distance is greater than the length of the runway, the jet cannot stop
Let's calculate the speed you should have to stop on a track of this size
Vo² = 2 a X
Vo = √ (2 7.42 800)
Vo = 109 m / s
It is conclusion the jet must lose some speed to land on this track
\[S = \Delta Q / T\] where S - entropy, Q heat and T temperature \[S = 295/402 = 0.733 \]
SO IT'S B
<h2>Hello!</h2>
The answer is: A. 19.3 joules
<h2>
Why?</h2>
Since it's an elastic collision, the kinetic energy after and before the collision will be the same.
Kinetic energy can be calculated using the following equation:
Where:
So,
First object, (going to the right):
Second object:, (going to the left):
Remember,
Hence,
The total kinetic energy after the collision will be:
The total kinetic energy after the collision is 19.3 joules (rounded to the nearest tenth)
Have a nice day!
