At that point it is no longer trying to uncompress nor is it trying to stretch. This is the same thing as a pendulum at the bottom of its swing, no longer falling but not yet rising against gravity. Thus the kinetic energy there is the same as the potential energy when it is compressed. The energy of compression is
This gives E=0.5(37)(0.2)²=
0.74JThis is the same as the kinetic energy when it is at natural length
Answer:
-30 °C
Explanation:
First, we have to calculate the molality (m) of the solution. If the solution is 50% C₂H₆O₂ by mass. It means that in 100 g of solution, the are 50 g of solute (C₂H₆O₂) and 50 g of solvent (water).
The molar mass of C₂H₆O₂ is 62.07 g/mol. The moles of solute are:
50 g × (1 mol / 62.07 g) = 0.81 mol
The mass of the solvent is 50 g = 0.050 kg.
The molality is:
m = 0.81 mol / 0.050 kg = 16 m
The freezing-point depression (ΔT) can be calculated using the following expression.
ΔT = Kf × m = (1.86 °C/m) × 16 m = 30 °C
where,
Kf: freezing-point constant
The normal freezing point for water is 0°C. The freezing point of the radiator fluid is:
0°C - 30°C = -30 °C
Answer:
In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.
<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
A) no H30+ ions or OH- ions.
