Answer:
She must stop the car before interception, distance traveled 12.66 m
Explanation:
We will take all units to the SI system
Vo = 48Km / h (1000m / 1Km) (1h / 3600s) = 13.33 m / s
V2 = 70 Km / h = 19.44 m / s
We calculate the distance traveled before stopping
X = Vo t + ½ to t²
Time is what it takes traffic light to turn red is t = 2.0 s
X = 13.33 2 + 1.2 (-7) 2²
X = 12.66 m
It stops car before reaching the traffic light turning to red
Let's analyze what happens if you accelerate, let's calculate the acceleration of the vehicle
V2 = Vo + a t2
a = (V2-Vo) / t2
a = (19.44-13.33) /6.6
a = 0.926 m / s2
This is the acceleration to try to pass the interception, now let's calculate the distance it travels in the time the traffic light changes from yellow to red (t = 2.0 s)
X = Vo t + ½ to t²
X = 13.33 2 + ½ 0.926 2²
X = 28.58 m
Since the vehicle was 30 m away, the interception does not happen
Trees. Every time the wind blows there is a wave of motion which is movement
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
