An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of o.01 s. It (A) must be zero (C) is n
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Answer:
The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.
Explanation:
Given;
wavelength of light, λ = 600 nm
The minimum thickness of the soap bubble for destructive interference to occur is given by;
where;
n is refractive index of soap film = 1.33
Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.
Answer : The partial pressure of is, 67.009 atm
Solution : Given,
Partial pressure of at equilibrium = 30.6 atm
Partial pressure of at equilibrium = 13.9 atm
Equilibrium constant =
The given balanced equilibrium reaction is,
The expression of will be,
Now put all the values of partial pressure, we get
Therefore, the partial pressure of is, 67.009 atm