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bixtya [17]
3 years ago
12

The pupil of an eagle’s eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a distance of 185 m, the eag

le sees them as one unresolved object and dives toward them at a speed of 16 m/s. Assume that the eagle’s eye detects light that has a wavelength of 550 nm in vacuum. How much time passes until the eagle sees the mice as separate objects?
Physics
1 answer:
max2010maxim [7]3 years ago
4 0

Answer:

time is 5.973826 sec

Explanation:

Given data

diameter D =  6.0 mm  6× 10^{-3} m

separated d =  0.010 m

distance (dis) 185 m

speed s = 16 m/s

wavelength = 550 nm = 550  10^{-9} m

to find out

How much time passes

solution

we know that for resolution  we use Rayleigh's Criterion i.e

θ = 1.22 wavelength  / diameter  =  separated / distance 1

we calculate distance 1 by put value wavelength, diameter and   separated

distance 1  = diameter × separated / 1.22 wavelength

distance 1  = 6× 10^{-3} × 0.010   / 1.22 × 550 × 10^{-9}

distance 1 = 89.418778

so time will be i.e = distance (dis) - distance 1  / speed

time = ( 185 -  89.418778) / 16

time = 5.973826 sec

time is 5.973826 sec

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4 0
2 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
2 years ago
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