How many revolutions does the object make during the first 3.9 s?

The area which is located under the curve gives us the angle in radians; we divide this by 2π to get units in revs (revolutions).

So for 0 ≤ t ≤ 2,

number of revs = 20rad/s * 2s * 1rev/2π rads = 6.37 revs 

So for 2 ≤ t ≤ 3.9,

number of revs = 10rad/s * 1.9s * 1rev/2πrads = 3.02 revs 

Therefore the total is about:

6.37 revs + 3.02 revs = 9.39 revs ≈ 9.4 revs

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.)

car a and b are moving in opposite directions

plz mark.me the brainliest i rlly need this kind of yours in exchange