Answer:
a) Q1= Q2= 11.75×10^-6Coulombs
b) Q1 =15×10^-6coulombs
Q2 = 38.75×10^-6coulombs
Explanation:
a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as
1/Ct = 1/C1 + 1/C2
Given C1 = 3.00 μF C2 = 7.75μF
1/Ct = 1/3+1/7.73
1/Ct = 0.333+ 0.129
1/Ct = 0.462
Ct = 1/0.462
Ct = 2.35μF
V = 5.00Volts
To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance
Q = 2.35×10^-6× 5
Q = 11.75×10^-6Coulombs
Since same charge flows through a series connected capacitors, therefore Q1= Q2=
11.75×10^-6Coulombs
b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2
C = 3.00 μF + 7.75 μF
C = 10.75 μF
For 3.00 μF capacitance, the charge on it will be Q1 = C1V
Q1 = 3×10^-6 × 5
Q1 =15×10^-6coulombs
For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5
Q2 = 38.75×10^-6coulombs
Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.
If you take a look at the definition of work, you will get this formula:
Answer:
10.337m/s2
Explanation:
F=ma
a=F/m
a = 92 / 8.9 = 10.337m/s2
Answer:
<h2> $1.50</h2>
Explanation:
Given data
power P= 2 kW
time t= 15 min to hours = 15/60= 1/4 h
cost of power consumption per kWh= 10 cent = $0.1
We are expected to compute the cost of operating the heater for 30 days
but let us computer the energy consumption for one day
Energy of heater for one day= 2* 1/4 = 0.5 kWh
the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50
<u><em>Hence it will cost $1.50 for 30 days operation</em></u>
