1. i drop a penny from the top of the tower at the front of fort collins high school and it takes 1.85 seconds to hit the ground. calculate the velocity in m/s after 1.10 seconds of freefall and calculate the velocity at impact in mi/hr.

The acceleration of gravity on Earth is  9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.

If you 'drop' the penny, then it starts out with zero speed. 
If you also start the clock at the same instant, then

         After  1.10 sec,  Speed = (1.10 x 9.8) = 10.78 meters/sec

         After  1.85 sec,  Speed = (1.85 x 9.8) = 18.13 meters/sec

But you want this second one given in a different unit of speed.
OK then:

     =  (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)

     =    (18.13 x 3,600 / 1609.344)  (mile/hr)  =  40.56 mph  (rounded)

We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'.  We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.

With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.

No bc a lot of things r still theories n unproven

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2   where t1=temperature of cold reservoir ( ° k) = 135            t2=temperature of hot reservoir ( ° k)   = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58%  note: this is the theoretical maximum efficiency of a perfect heat engine.   real-life engines do not achieve this value.