Aband-pass filter has fc1 = 5khz and fc2 = 88khz. calculate bandwidth (bw) and center frequency (fo) for this filter.

Bandwidth of the Band Pass Filter, BW = 88 - 5 = 83 kHz

center frequency, f = 20.976 kHz

Given:

Upper cut off frequency, = 88 kHz

Lower cut off frequency, = 5 kHz

Solution:

Bandwidth of the Band Pass Filter is the difference between its upper cut-off frequency and its lower cut-off frequency and thus is given by:

Bandwidth, BW = -

Bandwidth of the Band Pass Filter, BW = 88 - 5 = 83 kHz

Now, to calculate its center frequency, f:

f =

f =  

f = 20.976 kHz

the answer would most likely be c, because if it's a comical piece of information, it may likely not be true, if it's a site that allows anybody to edit the content, it could mean a five year old is typing the info, and if it's a tv show produced by a celeb, it may have a biased opinion.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft