1. Frequency: 
The frequency of a light wave is given by:

where
is the speed of light
is the wavelength of the wave
In this problem, we have light with wavelength

Substituting into the equation, we find the frequency:

2. Period: 
The period of a wave is equal to the reciprocal of the frequency:

The frequency of this light wave is
(found in the previous exercise), so the period is:

Answer:
P V = N R T ideal gas equation
P2 / P1 = T2 / T1 where the other variables are constant
P2 = (T2 / T1) * P1 = (313 / 293) * 40 psi = 42.7 psi
Answer:
a. 7.046 Nm²/C
b. 2.348 Nm²/C
Explanation:
Data given:
Base of equilateral triangle = 25.0 cm = 0.25 m
Strength of electric field = 260 N/C
In order to find the electric flux we first have to find out the area of triangle.
Area of triangle = 
= 
= 0.0271 m³
Lets find electric flux,
Electric Flux = E. A
= 260×0.0271
= 7.046 Nm²/C
Now we can find the electric flux through each of the three sides.
Electric flux through three sides = 
= 2.348 N m²/C
Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;
c) r>b E=ρ b (b-a)/r*εo
Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.
As we know,
∫E.dr= Qinside/εo
For r<a --->Qinside=0 then E=0
for a<r<b er have
E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L
E*2π*r*L =ρ*2*π*r* (r-a)*L/εo
E=ρ*(r-a)/εo
Finally for r>b
E*2π*r*L =ρ*2*π*b* (b-a)*L/εo
E=ρ*b* (b-a)*/r*εo