1) D
2) D.) Greater than 
Explanation:
1)
The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.
This phenomenon occurs only if two conditions are satisfied:
- The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
- The angle of incidence is greater than a certain angle called critical angle
In picture 1, we have 4 different diagrams. In the diagrams:
- The red arrow represents the incident ray
- The green arrow represents the refracted ray
- The blue arrow represents the reflected ray
Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)
2)
As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be
D.) Greater than 
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:
0.25 m/s
Explanation:
From the law of conservation of momentum
Mu+ mu = Mv' + m v
M= mass of the astronaut = 80 kg
m= mass of the oxygen tank= 10 Kg
v= speedof the tank 2 m/s
u= initial velocity of the system= 0
If we substitute the values, we have
( 80× 0 )+(10×0)= [(80 x v )+ (10 x 2)]
0= 80v + 20
-20=80v
v= -0.25 m/s ( we have a negative value because the astronaut and the motion of the cylinder are in opposite direction)
Hence the velocity the astronaut start to move off into space is 0.25 m/s