Answer:
The answer is A. Which is true
Answer:
well its simple the worker wouldn't actually be working
This behavior helps Betty in <u>intellectual </u>development.
Answer:
Explanation:
Given that;
horizontal circle at a rate of 2.33 revolutions per second
the magnetic field of the Earth is 0.500 gauss
the baton is 60.1 cm in length.
the magnetic field is oriented at 14.42°
we wil get the area due to rotation of radius of baton is

The formula for the induced emf is




B is the magnetic field strength
substitute


The magnetic field of the earth is oriented at 14.42

we plug in the values in the equation above
so, the induce EMF will be

