Answer:
The z-component of the force is
Explanation:
From the question we are told that
The charge on the particle is
The magnitude of the magnetic field is 
The velocity of the particle toward the x-direction is 
The velocity of the particle toward the y-direction is

The velocity of the particle toward the z-direction is

Generally the force on this particle is mathematically represented as

So we have

substituting values
So the z-component of the force is
Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).
Answer:
536,904 J/s
Explanation:
The energy output from motor is the input energy in the machine.
We know that efficiency is percentage energy ouput to energy input, and expressed as

Where n and E represent efficiency and energy respectively, subscripts o and i represent output and input respectively. Since for the machine we have the input energy then the output will be the product of efficiency and input energy
Energy output=0.6*1200 hp=720 hp
Converting hp to J/s we multiply by 745.7
Energy is 720*745.7=536,904 J/s
Disagree.
Fluoresce objects will only glow when put under actual Ultraviolet light. This is due to the molecules becoming excited by the ultraviolet radiation.
Microwaves give micro-waves that are present in another spectrum of wave length and will not be able to fluoresce the molecules. If it’s not “ultra violet “.... it’s not going to glow.
Answer:

Explanation:
As we know that there is no external torque on the system of two disc
then the angular momentum of the system will remains conserved
So we will have

now we have

also we have

now from above equation we have

now we have

