Answer:
Periodic.
Explanation:
Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.
Generally, the most commonly used electromagnetic wave technology in telecommunications is radio waves.
Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.
Basically, as a result of radio waves having long wavelengths, they are mainly used in long-distance communications such as the carriage and transmission of data.
Generally, a fixed speed is used for the propagation of traveling waves and this speed is usually denoted with the variable "v" or sometimes "c."
Furthermore, if the waveform of a traveling wave is repeated every time at specific intervals T, it is referred to as periodic wave.
Mathematically, the period of a traveling wave is given by the formula;
Where;
T is the time measured in seconds.
Answer:
In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.
Explanation
once the magnetic field weakens the area and cold plasma enters the area of the sunspot
N2(g)<span> + 3H</span>2(g)<span> → 2NH</span><span>3(g) Is the answer. </span>
Answer:
(a) 6650246.305 N/C
(b) 24150268.34 N/C
(c) 6408227.848 N/C
(d) 665024.6305 N/C
Explanation:
Given:
Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]
Total charge of the ring (Q) = 75.0 μC = [1 μC = 10⁻⁶ C]
Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:
Plug in the given values for each point and solve.
(a)
Given:
,
Electric field is given as:
(b)
Given:
,
Electric field is given as:
(c)
Given:
,
Electric field is given as:
(d)
Given:
,
Electric field is given as:
Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev