Answer:
B. q = 2.5 x 10^-1 C
Explanation:
Formula: E = F/q
Given information:
E = 40 N/C
F = 10N
--> q = F/E = 10/40 = 0.25 C
or q = 2.5 x 10^-1 C
Try to split the questions for easy access
Answer:
the volume in terms of time t ⇒ ![\\V = e^{0.05t} + 399\\](https://tex.z-dn.net/?f=%5C%5CV%20%3D%20e%5E%7B0.05t%7D%20%2B%20399%5C%5C)
Explanation:
Given that :
![\frac{dv}{dt}= kv\\\\\int\limits \, \frac{dv}{v}= \int\limits \, kdt\\In v = kt + C_1\\v = e^{kt} + C\\400 = e^{k*0} + C\\400 = 1 + C\\C = 400 -1\\C = 399](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%3D%20kv%5C%5C%5C%5C%5Cint%5Climits%20%5C%2C%20%5Cfrac%7Bdv%7D%7Bv%7D%3D%20%20%5Cint%5Climits%20%5C%2C%20kdt%5C%5CIn%20v%20%3D%20kt%20%2B%20C_1%5C%5Cv%20%3D%20e%5E%7Bkt%7D%20%2B%20C%5C%5C400%20%3D%20e%5E%7Bk%2A0%7D%20%2B%20C%5C%5C400%20%3D%201%20%2B%20C%5C%5CC%20%3D%20400%20-1%5C%5CC%20%3D%20399)
![V = e^{kt} + 399](https://tex.z-dn.net/?f=V%20%3D%20e%5E%7Bkt%7D%20%2B%20399)
When v = 300 ; ![\frac{dv}{dt}= - 15](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%3D%20-%2015)
then
![\frac{dv}{dt}= kv\\\\-15 = 300*k\\\\k = \frac{-15}{300}\\\\\\k = \frac{-1}{20}\\\\k = -0.05](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%3D%20kv%5C%5C%5C%5C-15%20%3D%20300%2Ak%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B-15%7D%7B300%7D%5C%5C%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B-1%7D%7B20%7D%5C%5C%5C%5Ck%20%3D%20-0.05)
∴ ![\\V = e^{0.05t} + 399\\](https://tex.z-dn.net/?f=%5C%5CV%20%3D%20e%5E%7B0.05t%7D%20%2B%20399%5C%5C)
Therefore, the volume in terms of time t ⇒ ![\\V = e^{0.05t} + 399\\](https://tex.z-dn.net/?f=%5C%5CV%20%3D%20e%5E%7B0.05t%7D%20%2B%20399%5C%5C)
Explanation:
a) 120m
b) since Q is the first order constructive interference, the distance between mid point of antennas and Q is 0.5 wavelengths (appear in A-Level question). so the wavelength should be (2)(30) =60m
60m