Work = (force) x (distance)
The work he did: Work = (700 N) x (4m) = 2,800 joules
The rate at which
he did it (power): Work/time = 2,800 joules/2 sec
= 1,400 joules/sec
= 1,400 watts
= 1.877... horsepower (rounded)
formed by the process of glaciation.
Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
