An alternating emf source with amplitude 14.9 14.9 V and an adjustable driving frequency f d fd is connected in series with a 59.4 59.4 Ω resistor and a 23.7 23.7 μF capacitor. At what value of f d fd are the amplitudes of the voltages across the resistor and capacitor equal?

113.0 Hz

Explanation:

In a RC circuit with alternating current, the voltage across the resistor is given by Ohm's Law:

where

I is the current in the circuit

R is the resistance of the resistor

While the voltage across the capacitor is given by:

where

is the impedance of the capacitor, where

f is the frequency

C is the capacitance

The voltages across the resistor and the capacitor are equal when

So

Which can be rewritten as

In this problem:

is the resistance

is the capacitance

Substituting, we find the frequency at which this happens:

  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

answerdmitry mendeleyev made a workable classification of the elements