Answer:
Explanation:
An information contains
25Hz and 75Hz sine wave
Sample frequency is 500Hz
The analogy signal are generally
y(t) = Asin(2πx/λ - wt), w=2πf
y1(t)=Asin(2πx/λ - wt)
y1(t)=Asin(2πx/λ - 2π•25t)
y1(t)=Asin(2πx/λ - 50πt)
Similarly
y2(t)=Asin(2πx/λ - 150πt)
Using Nyquist theorem
Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.
From sampling
f(nyquist)=f(sample)/2
f(nyquist)=500/2
f(nyquist)=250Hz
From signal
The highest frequency is 150Hz
F(nyquist) = 2×F(highest)
f(nyquist)= 2×150
f(nyquist)= 300Hz
Sample per frequency Ns is given as
Ns=F(sample)/F(highest signal)
Ns=500/150
Ns=3.33sample/period
This is above nyquist rate of 2sample/period
So signal below 300Hz reproduced without aliasing.
The highest resulting frequency is 300Hz
Wavelength - the distance from one wave crest or trough to another wave crest or trough. Amplitude - the distance from the median point or "middle" of the wave straight up to a crest (a maximum) or straight down to a trough (or minimum), which is the peak amplitude; or the distance from a trough straight up to a crest, or a crest straight down to a trough, called peak-to-peak amplitude.
Answer:
I wish I learned what I could do in the real world with the information I learned
Explanation:
The ability of the ground water to pass through the pore spaces in the rock is described as the rock's permeability. Permeable layers of rock that store and transport water are called aquifers.
Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;
solving this two equations together;
where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t
Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
