Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t
2 answers:
Answer:
final velocity = 0.08585m/s
Explanation:
We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.
The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after
total linear momentum before = linear momentum of first train car + linear momentum of second train car
We know that linear momentum = mv
where,
m = mass
v = velocity
thus,
total linear momentum before = m₁v₁ + m₂v₂
m₁ = mass of first train car = 135,000kg
v₁ = velocity of first train car = 0.305m/s
m₂ = mass of first second car = 100,000kg
v₂ = velocity of second train car = −0.210m/s
Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.
total linear momentum before = m₁v₁ + m₂v₂
= 135,000x0.305 + 100,000x(−0.210)
= 135,000x0.305 - 100,000x0.210
= 20,175 kgm/s
Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).
Therefore, total linear momentum after = mv
m = m₁ + m₂ = 135,000 + 100,000 = 235,000
total linear momentum before = total linear momentum after
235,000v = 20,175
v =
= 0.08585m/s
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1) Medium "b" has more optical density
2) Light must hit the interface between the two mediums perpendicularly
Explanation:
1)
Refraction occurs when light propagates from a medium into a second medium.
The optical density of a medium is given by its index of refraction, which is defined as:
where
c is the speed of light in a vacuum
v is the speed of light in a medium
Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.
In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.
2)
We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:
where
are the index of refraction of the two mediums
are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)
Here we want the direction of propagation of the light ray not to change: this means that it must be
(1)
However, here we have two mediums "a" and "b" with different index of refraction, so
Therefore the only angle that can satisfy eq.(1) is
So, the light must hit the surface perpendicular to the interface between the two mediums.
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Answer:
Plane will 741.6959 m apart after 1.7 hour
Explanation:
We have given time = 1.7 hr
So if we draw the vectors of a 2d graph we see that the difference in angles is =
Speed of first plane = 730 m/h
So distance traveled by first plane = 730×1.7 = 1241 m
Speed of second plane = 590 m/hr
So distance traveled by second plane = 590×1.7 = 1003 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.
Using the law of cosine, representing the distance between the planes, we see that:
r = 741.6959 m