Ii) The distance travelled by an object in time (100 + 1) s is (5.2 + 0.1) m. What is the
speed and it's error?
Ans: 0.052 ms-l, + 0.029 ms'

  v = (5.2 ± 0.2) 10⁻²   m / s

Explanation:

This is an exercise in kinematics of uniform motion. The formula for the average speed is

          v = x / t

let's calculate

          v = 5.2 / 100

          v = 0.052 m / s

          v = 5.2 10-2 m / s

we have to find the uncertainty of this quantity we propagate the error using derivatives

          Δv = dv /dx   Δx + dv /dt   Δt

taking everything in the most unfavorable case, I mean all positive

let's make the derivatives

          dv / dx = 1 / t

          dv / dt = -x / t²

we substitute

         Δv = 1 /t  Δx + x /t²   Δt

let's calculate the error

        Δv = 1/100 0.1 + 5.2 / 100²  1

        Δv = 1 10⁻³ + 5.2 10⁻⁴

        Δv = 1.52 10⁻³ m/s

the error must be given with a significant figure

         Dv = 2 10⁻³ m / s

the answer is

         v = (5.2 ± 0.2) 10⁻²   m / s

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j

answer///isobars;