A 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

The energy stored is

Explanation:

From the question we are told that  

   The capacitance  is  

    The resistance is  R = 3.00-Ω

    The emf is  

      The power  is  P = 300 W

Generally the total  emf is mathematically represented as

Here   is the emf across that capacitor which is mathematically represented as

and is the emf across the resistor which is mathematically represented as

So  

=>    

Generally the energy stored in a capacitor is mathematically represented as

=>      

=>      

=>      

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

answerscientific theory;