Answer:
Since the wire is not splitting at any point in the circuit,
the resistors are in series
Hence, Equivalent resistance = 10 + 20 + 30
Equivalent Resistance = 60 Ω
Answer: 200 J
Explanation: In order to explain this we have consider that the work done in a electric field is given by:
Work= Q*ΔV=2*100=200J
A.) kiloliter. 1 kiloliter = 1,000 liters
c.) megaliter. 1 megaliter = 1,000,000 liters
hope this helps
Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = 
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
<u>Determine the molar conductivity of acetic acid</u>
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
Answer:
Efficiency = 90 %
Wasted energy = 10 %
Explanation:
Since we have the input energy and useful output energy of the electric kettle, the only thing we are required to calculate here is its efficiency. This is gotten from
E = useful output energy/input energy × 100
E = 9/10 × 100 = 90 %
The percentage of wasted energy is
W = wasted energy/input energy × 100
W = 1/10 × 100
W = 10 %