At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. at this rate, how long (in seconds) does it take to accelerate from 91.0 km/hr to 104 km/hr?

t = 6.68 seconds

Explanation:

The acceleration of the automobile,

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

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.)