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.
The velocity of the stone when it is 5.25 m above the ground is determined as 6.93 m/s.
The velocity of the stone at the given displacement is calculated as follows;
K.E = ΔP.E
¹/₂mv² = mgh
v = √2gΔh
v = √[2(9.8)(7.7 - 5.25)]
v = 6.93 m/s
Thus, the velocity of the stone when it is 5.25 m above the ground is determined as 6.93 m/s.
Learn more about velocity here: brainly.com/question/4931057
#SPJ1
long and how
often have to do with exposure to chemicals?
Well how much, how long, and how often
For example how much is like how much a object is left in a small room with chemicals
How long is the time
how often is how much the same cycle is done repeatedly