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swat32
3 years ago
10

When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink

. To do this, the separation of the dots must be less than Raleigh's criterion. Take the pupil of the eye to be 3.2 mm and the distance from the paper to the eye of 42 cm; find the maximum separation (in cm) of two dots such that they cannot be resolved. (Assume the average wavelength of visible light is 550 nm.)
Physics
1 answer:
Dovator [93]3 years ago
4 0

Answer:

 y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

 Therefore the diffraction equation for slits with m = 1 remains

             a sin θ = λ

in general these experiments occur for oblique angles so

             sin θ = θ

             θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

           θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

          θ = s / R

as the angle is small the arc approaches the distance s = y

          y / R = 1.22 λ / s

          y = 1.22 λ R / a

let's calculate

            y = 1.22 500 10⁻⁹ 0.42 / 0.032

            y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

                 y <8 10⁻⁶ m

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Answer:

60 {ms}^{ - 2}

Explanation:

As we know,

=》Force = Mass × Acceleration

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=》Acceleration = 45 ÷ 0.75

=》Acceleration = 60

hence, the Acceleration of the ball would be. 60 meters per second square

60m {s}^{ - 2}

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3 years ago
How does newton's second law describe the motion of an object?
vodomira [7]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Explanation:

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3 years ago
Read 2 more answers
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn
kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (\vec{v}_{S}), measured in meters per second, is:

\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}

Where:

v_{o,x} - Initial velocity in +x direction, measured in meters per second.

a_{x} - Acceleration in +x direction, measured in meter per square second.

t - Time, measured in seconds.

v_{y} - Velocity in +y direction, measured in meters per second.

If we know that v_{o,x} = 0\,\frac{m}{s}, a_{x} = 0.250\,\frac{m}{s^{2}}, t = 45\,s and v_{y} = 21.4\,\frac{m}{s}, the final velocity of the satellite is:

\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}

\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}

\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }

\tan \alpha = 1.902

\alpha = \tan^{-1}1.902

\alpha \approx 62.266^{\circ}

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0
3 years ago
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stepan [7]

Answer:

25.5 m/s

Explanation:

The Doppler effect occurs when there is relative motion between a source of a wave and an observer. In such situation, there is a shift in the apparent frequency of the wave perceived by the observer.

The formula that gives the apparent frequency perceived by the observer is:

f'=\frac{v\pm v_o}{v\pm v_s}f

where

f is the real frequency of the wave

f' is the apparent frequency of the wave

v is the speed of the wave

v_s is the velocity of the source (negative if the source is moving towards the observer, positive otherwise)

v_o is the velocity of the observer (positive if the observer is moving towards the source, negative otherwise)

In this problem:

v = 340 m/s is the speed of sound

f = 800 Hz is the frequency of the horn

f' = 860 Hz is the apparent frequency

v_o=0 (the observer is at rest)

Re-arranging the equation for v_s, we can find the velocity of the horn and the driver:

f'=\frac{v}{v-v_s}f\\(v-v_s)f'=vf\\vf'-v_sf'=vf\\v_s=v\frac{f'-f}{f'}=(340)\frac{860-800}{860}=25.5 m/s

So, 25.5 m/s towards the observer.

3 0
3 years ago
Jake lifted a 10-kg bell by a distance of 0.50 meters in 0.50 seconds. How much power did Jake use?​
Reil [10]

Answer:

10 watts

Explanation:

first calculate work.

Work =force×distance cos thita

10Kg×0.50M cos 0= 5joules

Therefore, Power=Work÷ Time

Therefore, 5joules÷0.50s=10 watts.

8 0
2 years ago
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