Description : A lead sheet written in keyboard notation. Transcribed as performed on an album of the same name.
gdhfh
sacsFull description
Tutorial for learning Bentley Substation V8
Physical Chemistry, P Bahadur, IIT JEE, AIEEEFull description
Physical Chemistry, P Bahadur, IIT JEE, AIEEE
astrology
Physical Chemistry, P Bahadur, IIT JEE, AIEEEFull description
Summary of Brain Rules by John Medina
Full description
Q2.2 The resistance R resistance R(θ (θ ) of a thermostat at temperature θ K is given by by R R(θ (θ ) = α exp(β/θ ). Given that the resistance at the ice point (θ = 273.15 K) is 9.00 kΩ and the resistance at the steam point is 0.50 kΩ, find the resistance at 25 °C. Ans2.2 From the equation
R(θ R (θ ) = α exp(β/θ ).
the resistance at the ice point (θ = 273.15 K) is 9.00 kΩ 9 kΩ = α exp(β/273.15 ).
the resistance at the steam point is 0.50 kΩ
0.50 kΩ = α exp(β/373.15 ). To find the resistance at 25 °C we should be know the value of β and α
18 kΩ = exp(β/273.15 )/ exp(β/373.15 ) Now we use an technical numerically to find the value of β like the fixed point method . Then we will find β β = 2944.2 After that we use the value of β in the one of the above equation to find the value of α α = 9 kΩ / exp(2944.2/273.15 ) α = 1.86 × 10^−4
the resistance at 25 °C. R(298.15k) = (1.86 × 10^−4)* exp(2944.2/298.15 ) R(298.15k) = 3.62 kΩ = R( 25 25 °C )
