Answer:
Thats her fault.........................b
Explanation:
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
The spring scale will read 559 Newton's or 125.7 pounds.
Answer:
Electric field, E = 0.064 V/m
Explanation:
It is given that,
Resistivity of silver wire, 
Current density of the wire, 
We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :


E = 0.0636 V/m
or
E = 0.064 V/m
So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.
Could be easy for some people and hard for some people.