Since,
Speed = Frequency * WaveLength
=> WaveLength = Speed / Frequency --- (A)
Frequency = 13.0 kHz.
As the radio waves are electromagnetic waves, their speed is equals to the speed of light. Therefore,
Speed = C =
Plug in the values in equation(A):
A => WaveLength =
Ans: Wavelength = 23.077 kilometers.-i
Answer:
The Production Possibilities Curve (PPC) is a model used to show the tradeoffs associated with allocating resources between the production of two goods. The PPC can be used to illustrate the concepts of scarcity, opportunity cost, efficiency, inefficiency, economic growth, and contractions.
Explanation:
I hope this helps
Answer:
<h2>30 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question
force = 6 N
distance = 5 m
We have
workdone = 6 × 5 = 30
We have the final answer as
<h3>30 J</h3>
Hope this helps you
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
