Given two vectors A= 4.00i + 7.00jand B= 5.00i-2.00ja.Find the magnitude of each vector.b.Find the vector differenceC= A –B, giv
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Answer:
Explanation:
Part A) Using
light intensity I= P/A
A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2
Radius= Diameter/2
P= power= 10*10^-3=0.01 W
light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2
Part B) Using
I=c*ε*E^2/2
rearrange to solve for E= ((I*2)/(c*ε))
c is the speed of light which is 3*10^8 m/s^2
ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1
I= the already solved light intensity= 8.85*10^10 W/m^2
amplitude of the electric field E= (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)
---> E= (1.8*10^11) / (2.66*10^-3) =
(6.8*10^13) = 8.25*10^6 V/m
Answer:
The work done is "2000 J".
Explanation:
The given values are:
Force,
F = 200 N
Mass,
m = 55 kg
Displacement,
d = 10 m
Now,
The work done will be:
⇒
On substituting the given values, we get
⇒
⇒