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shutvik [7]
3 years ago
13

The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli

ght (wavelength = 500 nm)?
A. 0.01 rads
B. 10-5 rads
C. 10-3 rads
D. 10-4 rads
Physics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

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Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o
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Answer:

V = 20 miles /sec

Explanation:

We have remaining distance   =  d  = 96 miles

Lets call  Pascal velocity  V in miles per hour

Now if he increases his velocity by  50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V  = d/ t₁      ⇒   1.5 * V  =  96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 )     ⇒  V * (t₁ + 16 ) = 4*d     ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁      (1)

V*t₁  + 16*V = 384     (2)

We solve  from equation    (1)      t₁  = 64/V

And by substitution   in equation (2)

V * (64/V) + 16* V = 384

64  + 16 *V  = 384         ⇒   16*V = 320      ⇒  V= 320/16

V = 20 miles /sec

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3 years ago
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If the angle of refraction is 20 degrees what is the angle of incidence​
Romashka-Z-Leto [24]

Answer:

13.1

Explanation:

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5 0
3 years ago
On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8
xz_007 [3.2K]

Answer:

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density × volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π(15)³/3 = 14137.2 m³

Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg

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