Answer:
a, 1.775s
b, 17.04μC
c, 1.28s
Explanation:
Given
R = 1.25MΩ
C = 1.42µF
ε = 12.0 V
q = 8.78 µC
Time constant, τ = RC
τ = (1.25*10^6) * ( 1.42*10^-6)
τ = 1.775s
q• = εC
q• = 12 * 1.42*10^-6
q• = 17.04*10^-6C
q• = 17.04μC
Time t =
q = q• [1 - e^(t/τ)]
t = τIn[q•/(q•-q)]
t = 1.775In[17.04μC/(17.04μC-8.78μC)]
t = 1.775In(2.06)
t = 1.775*0.723
t = 1.28s
The range of frequencies of visible light in a vacuum is mathematically given as
Fmin=4.19*10^14Hz to Fmax=1*10^15Hz
<h3>What is the range of frequencies of visible light in a vacuum?</h3>
Question Parameters:
The wavelengths of visible light vary from about 300 nm to 700 nm.
Generally, the equation for the frequency is mathematically given as
F=C/\lambda
Therefore
For Fmax
Fmax=1*10^15Hz
Where
Fmin=4.19*10^14Hz
For more information on Wave
brainly.com/question/3004869
Answer:
what is the question? there isn't one
Answer:
Explanation:
Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .
Now the reflecting surface is twisted so that is becomes inclined at angle alpha .
The reflected light will be deviated from its original direction by angle
2 x alpha .
Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle
2 x beta
Hence angle between these two reflected beam
= 2 beta - 2 alpha
= 2 ( β - α )
So, angular separation between the rays reflected from the two surfaces
= 2 ( β - α ) .
Answer:
Hello your question is incomplete attached below is the complete question
answer : ri ( thickness of wire ) = 14.167 m
Explanation:
attached below is a detailed solution
using the given data to determine the thickness of wire
