One horsepower is about 745.7 watts
Answer:

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

Explanation:
For this case we know the mass of the water given :

And we know that the initial temperature for this water is
.
We want to cool this water to the human body temperature 
Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Where
represent the specific heat for the water and this value from tables we know that
for the water.
So then we have everything in order to replace into the formula of sensible heat and we got:

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

Answer:

Explanation:
Given:
- mass of rocket,

- time of observation,

- mass lost by the rocket by expulsion of air,

- velocity of air,

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)



They need more oxygen to function, so gas exchange needs to increase