C. Democritus was a Greek philosopher who defined the concept of atoms and is the only one of the answer choices who lived during that time period.
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
Answer:
We can solve this by analyzing the choices and the relations between variables.
Notice that Kinetic Energy depends on the speed and the mass. Actually the Kinetic Energy is directly proportional to those variables, which means the more mass or the more speed, the more Kinetic Energy would be.
Having said that, Part I and Part II must have speed and mass as independent variable, because Kinetic Energy depends on them.
Then, in the third question, the answer is Kinetic Energy because that's the dependent variable in both cases Part I and Part II.
Just to remember, the Kinetic Energy is defined as
Where you can notice the relation between mass, speed and kinetic energy.
Therefore, the right answer are Mass (Part I), Speed (Part II) and Kinetic Energy (Part III).
Explanation:
Answer:
the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
Explanation:
Given the data in the question;
slit separation d = 0.165 mm = 0.165 × 10⁻³ m
wavelength λ = 560 nm = 560 × 10⁻⁹ m
distance between the screen and slits D = 4.05 m
now,
for fifth-order bright fringe path difference = mλ
where m is 5
so, the difference in path lengths from each of the slits will be;
Δr = mλ
we substitute
Δr = 5( 560 × 10⁻⁹ m )
Δr = 28 × 10⁻⁷ m
Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m