Answer:
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) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
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
Answer:
The increase in the internal energy of the system is 360 Joules.
Explanation:
Given that,
Heat supplied to a system, Q = 292 J
Work done on the system by its surroundings, W = 68 J
We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.
Metals in general, are good heat conductors
Answer:
v= 335 m/s
2∆t= 0.75 s
∆x= v.∆t → ∆x= 335×½×0.75 = 125.625 m