The question here is how long does it take for a falling
person to reach the 90% of this terminal velocity. The computation is:
The terminal velocity vt fulfills v'=0. Therefore vt=g/c,
and so c=g/vt = 10/(100*1000/3600) = 36,000/100,000... /s. Incorporating the
differential equation shows that the time needed to reach velocity v is
t= ln [g / (g-c*v)] / c.
With v=.9 vt =.9 g/c,
t = ln [10] /c = 6.4 sec.
Answer:
Dear User,
Answer to your query is provided below
The time will be 8am on Wednesday when flight arrive in Tokyo.
Step-by-step explanation:
Here, You can see that when 11 a.m. in Johannesburg, it is 6 p.m. the same day in Tokyo, Japan ; this means 7Hrs. gap.
Further, A flight leaves Johannesburg at 6 a.m. on Tuesday for Tokyo takes 18 hours in the air, with an additional 1 hour stop-over on land.
So, 6a.m. + 18hr.+1hr.+7hr. = 8am on Wednesday in Tokyo (Arrival)
.bertha dived 8 times more then vernon 54/8 = 6
slope is change in y over change in x
use 2 points from the table so -3,-21 and -6,-39
change in Y: -39 - -21 = -18
change in x = -6 - -3 = -3
slope = -18/-3 = 6
slope = 6
