How much heat is released when 6.00 grams of ammonia and 5.00 grams of oxygen?
1 answer:
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Answer:
Visible light
X rays
ultraviolet radiation
gamma rays
microwave radiation
Explanation:
Electromagnetic waves consist of oscillating electric and magnetic fields which vibrate in a direction perpendicular to the direction of motion of the wave (transverse wave). Electromagnetic waves have all same speed in a vacuum (, known as speed of light) and are classified into 7 different types according to their frequency and wavelength. This classification is called electromagnetic spectrum.
From lowest to highest wavelength, the 7 types are:
Gamma rays
X-rays
Ultraviolet radiation
Visible light
Infrared radiation
Microwaves
Radio waves
Sound waves, on the contrary, do not belong to the electromagnetic spectrum, since they are another type of wave called mechanical waves (which consist of vibrations of the particles in a medium).
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
And therefore, the energy density at this distance from the center is
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
And therefore, the energy density at this distance from the center is