<span>You are given a circuit that contains a 6.0-v battery, a 4.0 ohm resistor, a 0.60 micro farad capacitor, an ammeter, and a switch all in series. You are asked to find the current reading after the switch is closed. Apply ohms law where V = IR where V is the voltage, I is the current and R is the resistor.</span>
V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A
When the switch is closed, the cathode side plate begins to fill up with electrons when it was originally empty before the switch was closed. When it fills up the cathode side of the circuit, the current decreases. And when the capacitor cannot hold more electrons, the current will stop. The higher the capacitance, the higher is the capacity to store electrons.
The closer the particles, the more will be the propogation of sound waves. Room contains air molecules which are far away from each other. So it takes much time for one molecule of air to disturb the other one. But in case of solids, as particles are much closer(compared to fluids), disturbance generated by one molecule is quickly transmitted to the next molecule
Answer:
Explanation:
The electrostatic potential energy for pair of charge is given by
U=1/4π∈₀×(q₁q₂/r)
Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:
U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)
U=3×1/4π∈₀×(q²/r)
Substitute given values
So
Answer:
Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. ... These rays of light will refract when they enter the lens and refract when they leave the lens.
Hope this helps...
