After 2.5 time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor C, charged through a resistance R?

impulse and momentum - a fire hose sends 20 kg of water each second onto a bearing building. If the water leaves the nozzle at 60m/s, what is the force exerted by the water on the building?

(II) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did each piece acquire?

To determine the percentage of the final voltage after 2.5 time constants, we need to understand the behavior of an RC circuit when charging a capacitor through a resistor.

The charging of a capacitor through a resistor follows an exponential growth curve according to the equation:

V(t) = V0 * [1 - exp(-t / (RC))]

where:
- V(t) is the voltage across the capacitor at time t
- V0 is the final voltage (EMF of the source)
- R is the resistance
- C is the capacitance
- exp is the exponential function with base e (Euler's number)
- t is the time

The time constant (τ) of the circuit is given by the product of the resistance and the capacitance (RC).

After one time constant (τ), the voltage across the capacitor will have reached approximately 63.2% of the final voltage (V0).

To find the percentage of the final voltage after 2.5 time constants, we substitute t = 2.5τ into the exponential growth equation:

V(2.5τ) = V0 * [1 - exp(-2.5)]

Now we can calculate the value of V(2.5τ) and express it as a percentage of the final voltage (V0).