A wave sent out from a source is reflected back in 1x10 -3 sec. If the velocity of the wave
is 3x 10²m/s what is the distance of the reflecting
Surface from the source​

The object's initial distance above the groundthe object's initial velocity

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.

Lexy is correct because angle r is between the incident wave,  n  = 0, and a diffracted wave.

'b ' is the correct conclusion.