Answer:
4000 Hz
Explanation:
An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.
The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.
Sampling rate = 2(max input frequency)
From the relation we can find out the cut-off frequency for the anti-aliasing filter.
max input frequency = sampling rate/2
max input frequency = 8100/2 = 4050 Hz
Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.
The correct answer is their difference in the charge to mass ratio a mass spectrometer, a compound is first vaporized and converted into ions .
Mass spectrometry is a technique for determining the mass-to-charge ratio (m/z) of one or more molecules in a sample. These measurements are frequently used to calculate the precise molecular weight of the sample components.
The speed at which positively charged ions move through a vacuum chamber toward a negatively charged plate is measured by mass spectrometers. ToF, magnetic sector, and quadrupole mass spectrometers are all commonly used in SIMS instrumentation.The MS/MS has three major advantages: the ability to study numerous molecules regardless of whether they are from the same structural family or not; the ability to highlight the specific metabolites of a disease; and it is an automated technique that allows for large-scale analysis.
Learn more about Mass spectrometry here :-
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Answer:
4.23 beats / seconds
Explanation:
Frequency heard = f = f₁ - f₂
f₂ = f₁ √ ( F₂ / F₁)
f = f₁ - f₁ √ ( F₂ / F₁) = f₁ ( 1 - √ ( F₂ / F₁) = 523 Hz ( 1 - 0.9919 ) = 4.23 beats / seconds
Answer:
in a microscope the place you keep your eyes to observe
Answer:
0.37sec
Explanation:
Period of oscillation of a simple pendulum of length L is:
T
=
2
π
×
√
(L
/g)
L=length of string 0.54m
g=acceleration due to gravity
T-period
T = 2 x 3.14 x √[0.54/9.8]
T = 1.47sec
An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.
The ball will first have V(max) at T/4,
=>V(max) = 1.47/4 = 0.37 sec
