Answer:
19 contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose.
Explanation:
Complete Question
Assume that the complete combustion of one mole of glucose to carbon dioxide and water liberates 2870 kJ/mol.
One contraction cycle in muscle requires 67 kJ, and the energy from the combustion of glucose is converted with an efficiency of 45% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.
Complete combustion of glucose liberates 2870 kJ/mol.
Complete combustion of one mole of glucose will liberate 2870 × 1 = 2870 kJ
The energy from the combustion of glucose is converted with an efficiency of 45% to contraction.
So, the amount of energy from the combustion of one mole of glucose that is converted to contraction is
45% × 2870 = 1,291.5 kJ
One contraction cycle requires 67 kJ of energy, so, 1291.5 kJ will cause
(1291.5/67) contraction cycles = 19.28 contraction cycles = 19 contraction cycles to the nearest whole number.
Hope this Helps!!!
Answer:
100J
Explanation:
Kinetic energy=1/2mv^2
Kinetic energy=(1/2 x 8)x5^2
Kinetic energy=4x25
Kinetic energy=100
100J
Answer:
Centre of mass of any body is a point where all mass of a body is supposed to be concentrated
it lies in geometrical centre....
Answer:
A=16.6 repeating
explanation:
take 100/6 and solve just make sure your equation is set up correctly!
and you should get an answer of 16.6 repeating or 166 m/hour or 16.6 meters/per/hour
Answer:
150000000
49050000 N/C
Explanation:
q = Charge = 24 pC
m = Mass of honeybee = 0.12 g
E = Electric field = 100 N/C
g = Acceleration due to gravity = 9.81 m/s²
Number electrons is
The number of electrons added or removed was 150000000
Force is given by
The ratio is
The ratio is
Balancing the forces we get
The electric field required is 49050000 N/C