Pretty sure it’s false....................
Answer:
Electromagnetic waves do not require a medium in order to transport their energy. Mechanical waves are waves that require a medium in order to transport their energy from one location to another. ... Sound is a mechanical wave and cannot travel through a vacuum.
Answer:
The electric field is 
Explanation:
Given that,
Radius = 2.00 cm
Number of turns per unit length 
Current 
We need to calculate the induced emf

Where, n = number of turns per unit length
A = area of cross section
=rate of current
Formula of electric field is defined as,

Where, r = radius
Put the value of emf in equation (I)
....(II)
We need to calculate the rate of current
....(III)
On differentiating equation (III)

Now, put the value of rate of current in equation (II)


Hence, The electric field is 
Answer:
The slope of a position-time graph can be calculated as:

where
is the increment in the y-variable
is the increment in the x-variable
We can verify that the slope of this graph is actually equal to the velocity. In fact:
corresponds to the change in position, so it is the displacement, 
corresponds to the change in time
, so the time interval
Therefore the slope of the graph is equal to

which corresponds to the definition of velocity.