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Zigmanuir [339]
3 years ago
7

you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh

ts that are 0.679 m apart. at what distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source? take the wavelength of the light to be 537 nm and your pupil diameter to be 4.81 mm.
Physics
1 answer:
inysia [295]3 years ago
7 0

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

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A 500 N force accelerates an object at 20 m s-2. What is its mass?
avanturin [10]

<u>Answer</u>: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

<u>Explanation:</u> Given force, F=500N

                                 acceleration, a=20 m/s^{2}

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

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                 ∴ Mass of the object is 25 kg .

<u> </u><u>Reference Link: </u>brainly.com/question/1141170

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8 0
2 years ago
Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,
Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

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The mass of a coin is measured to be 12.5±0.1 g. The diameter is 2.8±0.1 cm and the thickness 2.1 ±0.1 mm. Calculate the average
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The volume of the coin at the given diameter is calculated as follows;

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density = 9.67 g/cm³

Thus, the average density of the material from which the coin is made is 9.67 g/cm³.

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