I am not sure but i think the answer is C
Answer:

Explanation:
For the simple pendulum problem we need to remember that:
,
where
is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
,
where
is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
,
where T is the oscillation period. Now, we can easily solve for L:

Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
Answer:
false. it's positive terminal of an electric cell
I think it's false because Oxygen is diatomic so it's found as O2 which means that it has two atoms of itself.