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Levart [38]
3 years ago
9

To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 T

he pupil of the eye is the circular mm in dim light. Find the angular resolution of the eye for 550 nm opening through which light enters. wavelength light at those extremes. In which light can you see more sharply, dim or bright
Physics
1 answer:
Ronch [10]3 years ago
4 0

Correct question is;

To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright?

Answer:

We'll see more sharply in dim light

Explanation:

If we consider diffraction through a circular aperture, then angular resolution is given by;

θ = 1.22λ/D

where:

θ is the angular resolution (radians) λ is the wavelength of light

D is the diameter of the lens' aperture.

Thus,

at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm

θ = (1.22 * 550)/(2 x 10^(6))

θ = 335.5 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]

= 69.18 arc seconds

at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm

θ = (1.22 * 550)/(8 x 10^(6))

θ = 83.875 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]

= 17.3 arc seconds

From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.

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A 0.450 kg soccer ball has a kinetic energy of 119 J.
Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2  = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

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Sq rt v² = sq rt 528.89

V = 22.998m/s

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I hope this was helpful, please rate as brainliest

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3 years ago
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Anuta_ua [19.1K]

Answer:

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Explanation:

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8 0
2 years ago
If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict
Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol  × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT

R = 0.0821 L atm/ K mol  

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b =  0.03219 L / mol

So,  

\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15

P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}

P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}

⇒P  (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

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D. 6370.92 m

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