<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.
Boron, Aluminum, Gallium, Indium, Thallium
Old Grandpy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
741 J/kg°C
Explanation:
Given that
Initial temperature of glass, T(g) = 72° C
Specific heat capacity of glass, c(g) = 840 J/kg°C
Temperature of liquid, T(l)= 40° C
Final temperature, T(2) = 57° C
Specific heat capacity of the liquid, c(l) = ?
Using the relation
Heat gained by the liquid = Heat lost by the glass
m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)
Since their mass are the same, then
C(l)ΔT(l) = C(g)ΔT(g)
C(l) = C(g)ΔT(g) / ΔT(l)
C(l) = 840 * (72 - 57) / (57 - 40)
C(l) = 12600 / 17
C(l) = 741 J/kg°C
