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:
increasing; speeding up is my answer
Energy and Work have the same unit of measurement which is Joules in SI units.
Explanation:
- A Joule of Work is said to be done on an object when energy is transferred to that particular object.
- If two objects are involved, when one object transfers energy onto the second, a joule of work is said to be done by the first object.
- Work is also the application of force on an object over a distance. So Work = Force × Displacement
- Energy is neither created nor destroyed. It is in 2 forms - kinetic and potential.
- Kinetic energy is defined as the energy of a moving object while potential energy is known as the energy that is stored within an object.
- Kinetic Energy = 1/2 × mass × (velocity)²
- Potential Energy = mass × acceleration due to gravity × height
- Both energy and work are measured in Joules.
