A 14.0 gauge copper wire of diameter 1.628 carries a current of 12.0 mAmA . Part A What is the potential difference across a 2.25 mm length of the wire? VV = nothing VV SubmitRequest Answer Part B What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same?

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

a) The Potential difference across a 2.00 in length of a 14-gauge copper  

   wire:

  L= 2.00 m

From Table  Copper Resistivity = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

Explanation:

It is given that,

Initial gravitational potential energy of the ball-Earth-floor system is 10 J.

The ball then bounces back up to a height where the gravitational potential energy is 7 J.

Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.

Initial mechanical energy is 10 + 0 = 10 J

Mechanical energy just before the collision is 0 + 10 = 10 J

Final mechanical energy, 7 + 0 = 7 J

Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.