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
Isn't it "gravity" this would makes sense because grvaity difines weight
Protons, neutrons, and electrons<span> are the three main subatomic particles found in an atom. Protons have a</span>positive<span> (+) </span>charge<span>. An easy way to remember this is to remember that both proton and </span>positive<span> start with the</span>letter<span> "</span>P<span>." Neutrons have no electrical </span>charge<span>.</span>
Answer:
The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is
8.00 x 10-13J
Explanation:
In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).
Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)
Although, the kinetic energy is converted to potential energy in Coulomb's law equation.
That is,
1/2(mv^2) = (K* q1q2)/r
Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter
Answer:
The work done on the suitcase is, W = 600 J
Explanation:
Given,
The average force exerted by Jose on his suitcase, F = 60 N
Jose carried the suitcase to a distance, S = 10 m
The work done on the suitcase is given by the relation
<em>W = F x S</em>
Substituting the given values in the above equation,
W = 60 N x 10 m
= 600 J
Hence, the work done on the suitcase is, W = 600 J
