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.
b. Find the ball’s range from question 2.
Answer:
2.57 s
8.12 m
46.4 m
Explanation:
1. In the y direction:
s = 0 m
u = 22 sin 35° m/s = 12.6 m/s
a = -9.8 m/s²
Find: t
s = ut + ½ at²
0 = (12.6) t + ½ (-9.8) t²
t = 2.57 s
2. In the y direction:
u = 12.6 m/s
v = 0 m/s
a = -9.8 m/s²
Find: s
v² = u² + 2as
0² = (12.6)² + 2 (-9.8) s
s = 8.12 m
3. In the x direction:
u = 22 cos 35° m/s = 18.0 m/s
a = 0 m/s²
t = 2.57 s
Find: s
s = ut + ½ at²
s = (18.0) (2.57) + ½ (0) (2.57)²
s = 46.4 m
The angular speed in rpm will be 6. Angular speed is defined as the ratio of the revolution in the given time interval.
The rate of change of angulardisplacement is defined as angular speed.
ω = n/t
Where,
n is the number of revolution
t is the time
ω is the angular speed
ω = n/t
ω = 18/6
ω = 3 rpm
Hence,the angular speed in rpm will be 6 .
To learn more about the angular speed, refer to the link;
#SPJ2
b.forces
c.masses
d.gravity
Answer: B forces Hope this helps!!