Answer:
The 5 kHz signal will be less attenuated and the 10 kHz will be completely attenuated.
Low Pass Filter:
The purpose of a filter is to remove the undesired frequency components in a signal. A low-pass filter is one in which we allow low frequencies components to pass through it and block the higher frequency components. The frequency where it starts to attenuate the signal is called cut-off frequency.
Explanation:
We have a signal that contains 5 kHz and 10 kHz frequency component.
The cut-off frequency is 4 kHz which means the filter will attenuate the frequency components of 4 kHz or greater than that.
So, in theory both 5 kHz and 10 kHz frequency components will be attenuated.
But filters are not ideal and they usually have a little margin also called roll off which means it will allow some of the frequency components greater than cut-off frequency of 4 kHz.
Conclusion:
So to conclude, the 5 kHz signal will be less attenuated and the 10 kHz will be completely attenuated.
Answer:
The right answer is (d)
Explanation:
The pressure drop in smooth pipes with laminar flow is determined by the resistance of the fluid to flow, which is controlled by Reynolds's number. At higher Re, higher drop. But in this case, we have no information about the speed of the fluid in each pipe, and the ipe is not related to that speed. therefore we don´t have the information to say that any of the other options are the right ones
Answer:
The force acting on the bullet 
The value of impulse on the bullet in the given time interval P = 92.4 
Explanation:
Mass of the bullet ( m ) = 200 gr = 0.028 lb
Initial velocity ( U ) = 0
Final Velocity ( V ) = 3300 
Force acting on the bullet 
⇒ 
⇒
This is the force acting on the bullet.
Magnitude of the impulse imparted on the bullet 
-------- (1)
Put the value of F & dt in above equation we get,
P = 84000 × 0.0011
P = 92.4
This is the value of impulse on the bullet in the given time interval.
Answer:
15.64 MW
Explanation:
The computation of value of X that gives maximum profit is shown below:-
Profit = Revenue - Cost
= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2
= 14.7x - .47x^2 - 12
After solving the above equation we will get maximum differentiate for profit that is
14.7 - 0.94x = 0
So,
x = 15.64 MW
Therefore for computing the value of X that gives maximum profit we simply solve the above equation.
