Answer:
16.6 ms or 0.0166 s
Explanation:
If Q is the final charge, Q' is the initial charge, C the capacitance ,R is the resistance , t the time taken and τ the time constant,
[tex]Q = Q'( 1- e^{-t\div \tau })
τ = R C = (1.20×10³) (20×10⁻⁶) = 0.024 s
15 = 30 ( 1- e^{-t\div \ 0.024 })
( 1- e^{-t\div \ 0.024 }) = 15 ÷ 30
⇒ - e^{-t\div \ 0.024 }) = 0.5 -1
⇒ e^{-t\div \ 0.024 }) = 0.5
Taking logarithm to the base e on both sides of this equation,
⇒ t = 0.0166 seconds = 16.6 milli seconds
Answer:
d = 771.3m
Explanation:
Let's first calculate the time of flight:
where Voy=0. Solving for t:
Now we calculate the horizontal displacement, wihch is the distance from the target to drop the package:
Xf = d = Vox*t
d = 180*4.285 = 771.3m
Wave speed = frequency * wavelength
Wave speed = 686 * 2.00
Wave speed = 1,372 (m/s)
The answer to your question is A
Since they are both examples of moving waves, they both transmit energy.
