Answer : The wavelength of photon is, 
Explanation : Given,
Energy of photon = 
Formula used :

As, 
So, 
where,
= frequency of photon
h = Planck's constant = 
= wavelength of photon = ?
c = speed of light = 
Now put all the given values in the above formula, we get:


Conversion used : 
Therefore, the wavelength of photon is, 
Answer:
Hydrogen and helium compounds.
Explanation:
We know that the solar System was formed around <u>4.6 billion years ago, </u>due to the gravitational collapse of a giant interstellar molecular cloud.
This cloud is a type of interstellar cloud and its density and size permit the formation of molecules, most commonly molecular hydrogen.
Therefore the principal substances were found before planets began to form are hydrogen and helium compounds, besides Rocks, metals, most of them in gaseous form.
I hope it helps you!
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q =
is located in the center of a spherical cavity of radius ,
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)


Electric field at
m due to shell
E1 = 

Electric field at
due to 'q' at center 
E2 =

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)

Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² =
/ r
ω = √(
/ r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a
= g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849