E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
I think the correct answer is D: Potential Energy.
B) law of conservation of momentum
It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.
In this case total momentum before impact:
10kg*5m/s + 5kg * 0m/s = 50 kg m/s
After Impact:
10kg*0m/s + 5kg*10m/s = 50 kg m/s
You can see the momentum before and after impact is same as 50 kg m/s
Of course we assumed that the first cart stopped after the impact, and there are no energy losses.
1. I don't see a graph
2. It could be because a liquid could be evaporating in the process
3. It could be that the liquid is getting colder
4. 1 liquid is medium and moving a the normal pace
