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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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1 year ago
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

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b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

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Answer:

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Explanation:

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Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
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-- 'frictionless table';  don't need it.
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Answer:

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Explanation:

8 0
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