Answer:
c.
Explanation:
When you take samples of an analog signal, you must take samples fast enough so then you can recover the original signal, just passing the digitized signal through a low-pass filter.
The Nyquist criteria states that in order to be able to recover the original signal completely, you must take samples at a rate greater than 2 times the highest frequency component of the signal.
In our case, we should sample the analog signal at a rate > 2*5500 Hz = 11000 Hz.
If we sampled the 5500 Hz signal (assuming be a sinusoid just for simplicity) two times each cycle, we could be so unfortunate that the samples fall exactly when the signal crosses by zero, so the digitized signal would be only a train of zeros.
So, if we sample the signal only one time in each cycle, clearly we will not be able to recover the signal, and the digital signal will experience aliasing, due we are not following Nyquist sample rate requirements, as stated in the option c.
Answer:
h = 493 m
Explanation:
Given that,
A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 2.9 s later.
The total distance covered by the rock is 2d. We know that,
Speed = distance/time
So, the height of the cliff is equal to 493 m.
Answer:
80
Explanation:
I literally just did this question lol I hope I helped
Answer:
lanthanide and actinide
Explanation:
An inner transition metal (ITM) of chemical elements on the periodic table. They are normally shown in two rows below all the other elements. They include elements 57-71, or lanthanides, and 89-103, or actinides.
Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
