Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
Answer:
v = 2.45 m/s
Explanation:
first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height of cliff = 15 m
Vi = Initial Vertical Velocity = 0 m/s
t = time taken = ?
g = 9.8 m/s²
Therefore,
15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²
t² = (15 m)/(4.9 m/s²)
t = √3.06 s²
t = 1.75 s
Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,
s = vt
where,
s = horizontal distance covered = 4.3 m
v = original horizontal velocity = ?
Therefore,
4.3 m = v(1.75 s)
v = 4.3 m/1.75 s
<u>v = 2.45 m/s</u>
D or b you chose because it can go any way
