For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming
1 answer:
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Answer:
acceleration 8 km/h/s south
Explanation:
First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.
Based on this definition, we can already rule out the following two choices:
distance: 40 km
speed: 40 km/h
Since they only have magnitude, they are not vectors.
Then, the following option:
velocity: 5 km/h north
is wrong, because the car is moving south, not north.
So, the correct choice is
acceleration 8 km/h/s south
In fact, the acceleration can be calculated as
where
v = 40 km/h is the final velocity
u = 0 is the initial velocity
t = 5 s is the time
Substituting,
And since the sign is positive, the direction is the same as the velocity (south).
Answer:
The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°
Explanation:
According to Snell's Law,
n₁ sin θ₁ = n₂ sin θ₂
When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n
n sin θ₁ = n sin (90° - θ₁)
Note that from trigonometric relations,
Sin (90° - θ₁) = cos θ₁
n sin θ₁ = n cos θ₁
(sin θ₁)/(cos θ₁) = 1
tan θ₁ = 1
θ₁ = arctan 1 = 45°
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