3) A 100m track is marked off in meters. When a sprinter leaves the starting line, timers are started. The sprinter passes the 12m [E] mark at 1.8s and passes the 56m [E] mark at 6.7s. What was the sprinter’s velocity between those two positions?

Explanation:

A track 100m

The sprinter passes 12m mark in 1.8seconds

Let this be the initial point

d1 = 12m

t1 = 1.8seconds

And Passes the 56m mark in 6.7 seconds

Then, the second point is

d2 = 56m

t2 = 6.7 seconds

The sprinter velocity between this two position

Velocity is the rate of change of displacement

V = displacement / time taken

V = ∆x / ∆t

V = (x2 - x1) / (t2 - t1)

V = (56 - 12) / (6.7 - 1.8)

V = 44 / 4.9

V = 8.98 m/s

Then, the sprinter velocity between this two position is 8.98 m/s

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.)

568.98

explanation:

its math