A0.2 cm diameter wire must carry a 20-a current. if the maximum power dissipation along the wire is 4w/m, what is the minimum allowable conductivity of the wire in ohm-m? (a) 3.18x10 (b) 3.18x10 () 3.18x10 (d) 3.18x10

The conductivity of the wire is .

Explanation:

Given that,

Diameter = 0.2 cm

Current = 20 A

Power = 4 W/m

We need to calculate the conductivity

We know that,

Using  formula of resistance

....(I)

Where,

= resistivity

A = area

l = length

Using formula of power

Put the value of R in equation (I)

Put the all values into the formula

Hence, The conductivity of the wire is .

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

i think the answer is a, hope this : )