Answer:

Explanation:
<u>Given:</u>
- Length of the race = 10 km
- Distance traveled by Steve in 25 min = 7.57 km
- Time interval for the constant acceleration = 60 s
- Initial velocity of Steve = 0 m/s
<u>Assume:</u>
- u = initial velocity
- v = final velocity
- t = time interval
- a = constant acceleration
Since the person has the instantaneous velocity at 25 min equivalent to the average velocity of the person during this time interval. So, let us find out the average velocity of the person till 25 min time interval.

Since the person moves with a constant acceleration for the first 60 s and then moves with a constant velocity after this instant. This means the final velocity of the person at the end of 60 s is 5.047 m/s.

Now using the equation of constant acceleration, we have

Hence, the acceleration of Steve in the 60 s interval is
.
100 g lead target. Lead has a very high density which means even a very small volume and weight quite a lot. So because the lead target weighs only 100 g it's going to be small in size compared to other targets which are all made of lesser dense materials than lead.
Answer:
the answer for this is my.
To complete a task in a short amount of time/get from A to B in the quickest amount of time
Hope helps!
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:
