Answer:
<em>b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.</em>
<em></em>
Explanation:
The universe is thought to be made up of 85% dark matters. <em>Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect. This means that option b is wrong since radio wave is an electromagnetic wave</em>. Dark matter is a form of matter that makes up about a quarter of the total mass–energy density of the universe. Dark matter was theorized due a variety of astrophysical observations and gravitational effects that cannot be explained by accepted theories of gravity unless there were more matter in the universe than can be seen.
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He
Answer:
Explanation:
Given that,
Wavelength, 
We need to find the frequency of the violet light.
We know that the relation between frequency and wavelength is given by :
So, the frequency of violet light is 
.
