Answer:
Explanation:
Given
Charge of first Particle 
Charge of second Particle 
distance between them 

magnetic field due to first charge at mid-way between two charged particles is



(away from it)
Electric field due to 


(towards it)

(away from first charge)
To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.
The altitude is,

And the velocity can be written as,


From the properties of standard atmosphere at altitude z = 20km temperature is



Velocity of sound at this altitude is



Then the Mach number



So front stagnation temperature



Therefore the temperature at its front stagnation point is 689.87K
Answer:
Constructive Interference
Explanation:
Constructive Interference occurs when two waves superimpose and make bigger amplitudes.
In constructive interference, the crests of one wave fall on the crests of second wave and the amplitudes add up. The amplitude of the resultant wave is equal to sum of the amplitude of the individual waves. Similarly, the trough of first wave falls on the trough of other wave and they superimpose to create the trough of the resultant wave.
For Example, In the attachment, two waves A and B superimpose and demonstrate Constructive interference to create the wave C.
Answer:
The resonant frequency of this circuit is 1190.91 Hz.
Explanation:
Given that,
Inductance, 
Resistance, R = 150 ohms
Capacitance, 
At resonance, the capacitive reactance is equal to the inductive reactance such that,

f is the resonant frequency of this circuit



So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.
Answer:
what is your question because I seem to not see a question in this question....