The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.
Explanation:
Let at A both net electric field is zero then
At A ,E1=E2
E1=k*Iq1I / (d+x)^2
E2=k*Iq2I /x^2
Equating both
Answer:
Explanation:
This problem relates to interference of light in thin films .
The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.
2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .
2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.
2 x 1.5 t = 600 / 2 nm
t = 100 nm .
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ