No, that's silly.
You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level. All of those transitions ploop out
photons at Infrared wavelengths.
THEN next you get your Balmer series, where the electrons fall in
to the 2nd level. Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.
And then there's the Lyman series, where electrons fall all the way
down to the #1 level. Those are ALL in the ultraviolet.
Answer:
Incomplete question
This is the completed question
If the resistor in the circuit had a larger resistance then the current would be then have to be proportionally smaller. Because the batteries each give off 1.5 volts then the current would have to be the variable that would change. What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the current? What would happen to the resistor?
Explanation:
Using ohms law as our basis
Ohms law state that, the voltage in an ohmic conductor is directly proportional to the current
V∝I
Resistance is the constant of proportionality
Then
V=iR
Since we want a relationship between current and resistance.
then, I=V/R
So, current is inversely proportional to Resistance
as the current increase the resistance reduce and as the current reduces the resistance increases.
a. So, increasing the voltage from 1.5V to 12V increases the current In the circuit because voltage Is directly proportional to I.
From ohms law
V=iR
When v =1.5V
I=1.5/R
When V increase to 12V
I=12/R
I.e, it increases by a factor of 8. Eight times it's initial value
b. Now, the resistance in the circuit is the constant of proportionality and it doesn't change in a given circuit expect when using a variable resistoa r like rheostat.
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = 
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
It brings together combinations of product
Heat is an energy that is produced from something like the sun. Sound energy is when an object works to make sound. Light is when something gives off light by working. And potential energy is when an object is completely still but can be moved by another force
