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.
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Answer:
Explanation:
Check attachment for solution
Answer:
0.25 gram of neptunium is remaining
Explanation:
First we calculate the no. of half lives passed. For that we have formula:
n = t/T
where,
n = no. of half lives passed = ?
t = total time passed = 8 days (From Monday noon to Tuesday noon of following week)
T = Half Life Period = 2 days
Therefore,
n = 8 days/2 days
n = 4
Now, for the remaining mass of neptunium, we use the formula:
m = (mi)/(2)^n
where,
mi = initial mass of neptunium = 4.00 grams
m = remaining mass of neptunium = ?
Therefore,
m = 4 grams/2⁴
<u>m = 0.25 gram</u>
