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agasfer [191]
3 years ago
5

You are using a small telescope to view two closely spaced light bulbs located a mile away. Each light bulb is emitting blue lig

ht. At this distance, the light bulbs are still too close together for you to distinguish them as separate bulbs. If they were emitting red light instead, would you have a better or worse chance of distinguishing them as separate bulbs?
Physics
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

Explanation:

The problem is based on resolving power of instruments . Resolving power of instruments is related to wavelength of light as follows

resolving power of telescope ∝  1 / λ

Higher the resolving power closer the object that can be looked separately.

Wave length of blue light is shorter than that of red light so resolving power in case of blue light will be higher. So we have better chance of distinguishing separate bulbs when color light is blue.

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If a driver is tired, the thinking distance will be less. True or false.why?
Evgesh-ka [11]

Answer:

False

Explanation:

When a driver is impaired the brain will need an extra second or two to wake up, after it does wake up it tends to drift off regularly.

8 0
3 years ago
3 What is the displacement of a satellite when it makes a complete round along its circular path?​
kirill [66]

Answer:

0

Explanation:

The displacement is zero since it goes in a full circle and ends up where it started.

4 0
2 years ago
You see lightning before you hear thunder because A) light waves can travel through a vacuum. B) sound waves cannot travel throu
Alex Ar [27]
C) light waves travel faster than sound waves

You were correct
3 0
4 years ago
Read 2 more answers
g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a
zimovet [89]

Answer:

 λ = 3 10⁻⁷ m,   UV laser

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

let's use trigonometry

         tan θ = y / L

as in this phenomenon the angles are small

        tan θ = \frac{sin \ \theta}{cos \ \theta} = sin θ

        sin θ = y / L

we substitute

      a y / L = m  λ

let's apply this equation to the initial data

       a  0.04 / L = 1 600 10⁻⁹

       a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

      a 0.04 / L = 2  λ

       a / L = 50  λ

we solve the two expression is

         1.5 10⁻⁵ = 50  λ

          λ = 1.5 10⁻⁵ / 50

          λ = 3 10⁻⁷ m

    UV laser

3 0
3 years ago
How is work defined in physics
Tcecarenko [31]

Explanation:

Work is the dot product of the force and displacement vectors.

W = F · d

In other words, it is the force times the parallel component of the distance.

W = F d cos θ, where θ is the angle between the force and distance.

3 0
3 years ago
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