Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Answer:
5. -24 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s².
mathematically,
a = dv/dt ............................ Equation 1
Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.
But
v = dx(t)/dt
Where,
x(t) = 27t-4.0t³...................... Equation 2
Therefore, differentiating equation 2 with respect to time.
v = dx(t)/dt = 27-12t²............. Equation 3.
Also differentiating equation 3 with respect to time,
a = dv/dt = -24t
a = -24t .................... Equation 4
from the question,
At the end of 1.0 s,
a = -24(1)
a = -24 m/s².
Thus the acceleration = -24 m/s²
The right option is 5. -24 m/s²
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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