Acceleration = change in velocity/change in time
= (30 - 20) / 10 - 0
= 10 / 10
Acceleration = 1 m/s²
Answer: Hence the energy change is -1918 kJ/mol
Given:
Initial temperature of calorimeter T1 = 25 C
Final temperature T2 = 29 C
Heat capacity of calorimeter c = 4.90 kJ/C
Mass of sucrose (m) = 3.5 g
Molar mass of sucrose (M) = 342.3 g/mol
To determine:
The energy change for the combustion of sucrose in kJ/mol
Explanation:
The change in temperature of calorimeter = T2 -T1 = 29 - 25 = 4 C
Heat gained by the calorimeter when the temperature changes by 4 C is-
= 4 C * 4.90 kJ/ 1 C = 19. 6 kJ
Now,
Heat gained by calorimeter = heat lost during combustion of 3.5 g sucrose = -19.6 kJ
# moles of sucrose = 3.5 g/342.3 g.mole-1 = 0.01022 moles
Energy change for the combustion reaction = -19.6/0.01022 = -1917.8 kJ/mol
The voltage in the secondary coil is 20 V
Explanation:
The transformer equation gives the relationship between the voltage in the primary and secondary coil of a transformer:
where
is the voltage in the primary coil
is the voltage in the secondary coil
is the number of turns in the primary coil
is the number of turns in the secondary coil
It can also be rewritten as
In this problem, we have a step-up voltage that doubles the voltage, so:
The voltage in the primary coil is 10 V,
Therefore, the voltage in the secondary coil is
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