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Andrej [43]
2 years ago
14

Does every light source emit only one type of light?

Physics
1 answer:
blsea [12.9K]2 years ago
3 0

Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.

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Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove  = 300m

Time taken  = 18s

Unknown:

Speed  = ?

Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

   Speed  = \frac{distance}{time}

Insert the parameters and solve;

  Speed  = \frac{300}{18}  = 16.67m/s

3 0
3 years ago
What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?
Vikentia [17]

Explanation:

Momentum = mass × Velocity

p = 62×73

p =4526

4 0
3 years ago
What can we say about Gamma waves on<br> the electromagnetic spectrum?
sladkih [1.3K]

Answer: Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions. Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

Explanation: :)

6 0
2 years ago
1. The distance between a trough and a crest on a transverse wave is 12.0cm. If cycles of the wave pass a fixed point in one sec
11Alexandr11 [23.1K]

Answer:

0.12m/s

Explanation:

v=λf

Given that, λ = 12cm = 0.12m

T = 1second

(A period T is the time required for one complete cycle of vibration to pass a given point)

frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz

therefore, v= 0.12 × 1 = 0.12m/s

8 0
2 years ago
Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at
atroni [7]

 Answer:

y = 52.44 10⁻⁶  m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

         a sin θ= m λ

The first minimum occurs for m = 1

       sin θ =  λ  / a

Now let's use trigonometry the object is a distance L = 0.205 m

        tan θ = y / L

Since the angles are very small, let's approximate

        tan θ = sin θ/cos θ = sin  θ

        sin θ = y / L

We substitute in the diffraction equation

         y / L =  λ  / a

         y =  λ  L / a

Let's calculate

        y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

        y = 52.44 10⁻⁶ m

7 0
4 years ago
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