Answer:
The work required to cause this volume change is 2 x 10⁶ J
Explanation:
Given;
constant pressure of the gas, P = 100 kPa = 100,000 Pa
change in volume of the gas, ΔV = 20 m³
The work required to cause this volume change is calculated as;
W = PΔV
Substitute the given values and solve for the required work (W).
W = (100,000)(20)
W = 2 x 10⁶ J
Therefore, the work required to cause this volume change is 2 x 10⁶ J
Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
(1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC
Answer:
look in internet u can find online converter and there u will get a rel answer
Explanation: