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Zigmanuir [339]
4 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]4 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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An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

7 0
3 years ago
Please answer me!!! asappp​
Anastasy [175]

Answer:

omg this is hard

Explanation:

Yes it is lol

5 0
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A car is running at a velocity of 50 miles/hour and the driver accelerates the car by 10miles/hour.How far the car travels from
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Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
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