A - the objects are too small
GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf
Answer:
the answer would be "using more heat" btw
Explanation:
As per the question the mass of the boy is 50 kg.
The boy sits on a chair.
We are asked to calculate the force exerted by the boy on the chair at sea level.
The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.
At sea level the acceleration due to gravity g = 9.8 m/s^2
Hence the weight of the boy 
[m is the mass of the body]
we have m = 50 kg.
Hence w = 50 kg ×9.8 m/s^2
=490 N kg m/s^2
= 490 N
Here newton [N] is the unit of force.
A) use v=u+at for both
First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.
b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.
First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
The moon, because the acceleration due to gravity is less.
