Avery long, straight wire has charge per unit length 3.50×10^−10 c/m . at what distance from the wire is the electricfield magnitude equal to 2.90 n/c ?

r= 2.17 m

Explanation:

Conceptual Analysis:

The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:

E= 2k*(λ/r) Formula (1)

Where:

E: electric field .( N/C)

k: Coulomb electric constant. (N*m²/C²)

λ: linear charge density. (C/m)

r : distance from the charge line to the surface where E calculates (m)

Known data

E= 2.9  N/C

λ = 3.5*10⁻¹⁰ C/m

k= 8.99 *10⁹ N*m²/C²

Problem development

We replace data in the formula (1):

E= 2*k*(λ/r)

2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)

r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)

r= 2.17 m

second mirror gathers light from a faraway object and reflects it to another mirror. - a.

  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