Im pretty sure its C- Wavelength
Because the number of valence electrons of an element determines the properties and in particular the reactivity of that element.
In fact, elements of the first group (i.e. only one valence electron) have high reactivity, because they can easily give away their valence electron to atoms of other elements forming bonds. On the contrary, elements of the 8th group (noble gases) have their outermost shell completely filled with electrons, so they do not have valence electrons, and they have little or no reactivity at all.
Answer:
β = 114 db
Explanation:
The intensity of sound in decibles is
β = 10 log 
in most cases Io is the hearing threshold 1 10-12 W / cm²
let's calculate the intensity of each instrument
I / I₀ = 10 (β / 10)
I = I₀ 10 (β / 10)
trumpet
I1 = 1 10⁻¹² 10 (94/10)
I1 = 2.51 10⁻³ / cm²
Thrombus
I2 = 1 10⁻¹² 10 (107/10)
I2 = 5.01 10-2 W / cm²
low
I3 =1 1-12 (113/10) W/cm²
I3 = 1,995 10-1 W / cm²
when we place the three instruments together their sounds reinforce
I_total = I₁ + I₂ + I₃
I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1
I_total = 0.00251 + 0.0501 + 0.1995
I_total = 0.25211 W / cm²
let's bring this amount to the SI system
β = 10 log (0.25211 / 1 10⁻¹²)
β = 114 db
Answer:
a
The number of fringe is z = 3 fringes
b
The ratio is 
Explanation:
a
From the question we are told that
The wavelength is 
The distance between the slit is 
The width of the slit is 
let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

Substituting values
z = 3 fringes
b
From the question we are told that the order of the bright fringe is n = 3
Generally the intensity of a pattern is mathematically represented as
![I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%5B%5Cfrac%7B%5Cpi%20d%20sin%20%5Ctheta%7D%7B%5Clambda%7D%20%5D%5B%5Cfrac%7Bsin%20%28%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%20%7D%20%29%7D%7B%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%7D%20%7D%20%5D)
Where
is the intensity of the central fringe
And Generally 
![I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20co%5E2%20%5B%20%5Cfrac%7B%5Cpi%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%5D%20%5B%5Cfrac%7B%5Cfrac%7Bsin%20%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%7D%7B%5Cfrac%7B%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%28n%20%5Cpi%29%5B%5Cfrac%7B%5Cfrac%7Bsin%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%29%7D%7B%20%5Cfrac%7B%20%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%20%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%283%20%5Cpi%29%20%5B%5Cfrac%7Bsin%20%28%5Cfrac%7B3%20%5Cpi%20%7D%7B6%7D%20%29%7D%7B%5Cfrac%7B3%20%5Cpi%7D%7B6%7D%20%7D%20%5D)


neutron star is formed when the core of a star, that was one ginormous, collapse on its own mass thus it loses it's volume exponentially (y=a^x) while the mass of the star decreases linearly (y=mx+c) thus the neutron star has mass of 10 to 29 sun in the radius of 10 km( for scale earth has 6731 km) thus it's density is super high!!!!!!!!!!!!!!!!!!!
hope it helps you! ;)