Two long parallel wires carry currents of 3.05 A 3.05 A and 6.69 A 6.69 A . The magnitude of the force per unit length acting on each wire is 7.33 × 10 − 5 N / m 7.33×10−5 N/m . Find the separation distance d d of the wires expressed in millimeters.

The separation between wires is 55.6 mm.

Explanation:

Given that,

Current in wire 1,

Current in wire 2,

The magnitude of the force per unit length acting on each wire is,

We know that the formula of magnitude of the force per unit length is given by :

Here, d is the separation between wires.

So, the separation between wires is 55.6 mm.

  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

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