1. You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you at 44.2 m/s from behind and you perceive the frequency as 1310 Hz. You are relieved that he is in pursuit of a different speeder when he continues past you, but now you perceive the frequency as 1240 Hz. What is the frequency of the siren in the police car

The frequency of the siren in the police car is 1270.45 Hz

Explanation:

Here we have

f(v-35) = 1310(v-44.2)

fv -35f = 1310v - 57902....(1)

Also

f(v + 35) = 1240(v + 44.2)

fv + 35f = 1240v + 54808.....(2)

Subtracting (1) from (2) gives

70f = 112710 -70v

Therefore f = 1610.143 -v

Substituting the value of f in (1) we get

(1610.143 -v)v -35(1610.143 -v) = 1310·v - 57902

Which gives

v²-335.143·v-1546 =0

Factorizing gives

(v + 4.554)(v - 339.697) =0

Therefore, since v is the velocity of sound, we have v = 339.697 m/s

Since our f = 1610.143 -v

f = 1610.143 -339.697 = 1270.45 Hz

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

it is the second option.