Products such as antifreeze are composed of organic compounds that are classified as <em>alcohols</em>. (a)
Maybe those other classes of chemicals also lower the freezing temperature of water, just like alcohol does. I don't know. But alcohol is what's used to make anti-freeze. I'm guessing alcohol must be cheaper, less toxic, and less corrosive inside the engines' cooling systems than any of that other stuff is.
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

Recall that the speed of light is equivalent to

Then replacing,


Our values are given as




Replacing we have,



Therefore the magnetic field around this circular area is 
B. is the answer.
C is not correct because the light is actually reflected off of an opaque object.
Answer:
The diagram assigned B
explanation:
Check the direction of the two vectors, their resultant must be in the same direction.
To find average speed, we divide the distance of travel (in this case, 400 metres) by the time she took, 32 seconds. Therefore: 12.5 seconds is her average speed.