Where is the diagram? What is the question?
Answer:
Number of turns per metre, n= 500/0.3= (5000/3)m^-1
Cross sectional areaof the square loop of wire, A= (0.1^2)m^2= 0.01m^2dB/dt= μn(dI/dt)= (4.00π x10^-7)(5000/3)(0.7)= 1.46608x10^-3T/s
The induced emf in the square loop of wire, ε= the rate of change of magnetic flux of the square loop of wire(dΦ/dt)= A(dB/dt)= (0.01)(1.46608x10^-3)= 0.0146608x10^-5VA
current flows in the square loop of wire since a potential difference(induced emf in this case) exists. Its magnitude,
I= ε/R where R is the resistance of the square loop of wire.
I= (0.0146608x10^-5)/30= 4.89x10^-7A
Answer:
We are given x= bt +ct²
So
A. bxt= m
Because m/s*s= m
So b= m/s and c= m/s²
B.
x= bt-ct²
So at x=0 t=0
x=0 t= 2
We have
bt = ct² so t = b/c at x= 0
So b-2ct= 0
B. To find velocity we use
dx / dt = b - 2 Ct
C. At rest wen V= 0
We have t= b/2c
D. To find acceleration we use
dv / dt = - 2C
Answer:
Intensity of beam 18 feet below the surface is about 0.02%
Explanation:
Using Lambert's law
Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium
then dI / I = kdt
taking log,
ln(I) = kt + ln C
I = Ce^kt
t=0=>I=I(0)=>C=I(0)
I = I(0)e^kt
t=3 & I=0.25I(0)=>0.25=e^3k
k = ln(0.25)/3
k = -1.386/3
k = -0.4621
I = I(0)e^(-0.4621t)
I(18) = I(0)e^(-0.4621*18)
I(18) = 0.00024413I(0)
Intensity of beam 18 feet below the surface is about 0.2%
