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:
Hello!
After reading your question I have calculated the correct answer to be:
47.22
Step-by-step explanation:
The way I solved this equation was I first split the equation into two parts.
I took the two whole numbers, being "34" and "12", and made their own equation:
"34+12=?"
And then I took the two partial numbers "0.98" and "0.24" and made their own equation:
"0.98+0.24=?"
I then solved both equations:
"34+12= 46"
"0.98+0.24= 1.22"
And added those two solutions together to come up with the final solution:
"46+1.22= 47.22"
Hope this helped!
1 and 8 are because they are alternate exterior angles
Answer:
Hi can you please tell of which grade question is this ?
Answer:
it is D hope this helps
Step-by-step explanation:
