Answer:
The height is "89.61 m". A further explanation is given below.
Explanation:
According to the question,
Mass of rocket,
m = 5 kg
Time,
t = 4.6 s
Initial speed of rod
u = o m/s
Final speed,
v = 39 m/s
drag,
= 60 N
So,
⇒ acceleration, 


⇒ 



Now,
The thrust will be:
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
The height will be:
⇒ 



Answer:
Democritus
Explanation:
Democritus was a Greek philosopher who was the first person to use the term atom . However ,John Dalton was the first to adapt Democritus’ theory into the first modern atomic
model
In short, and in general:
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Answer:
don't know what class are you you are using which mobile or laptop
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:


We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:

And the velocity achieved in part 1:

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

Then, to get
, we do:



And we substitute the values:
