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.
16. 5x^3 y^-5 • 4xy^3
20x^4y^-2
20x^4 • 1/y^2
=20x^4/y^2
17. -2b^3c • 4b^2c^2
= -8b^5c^3
18. a^3n^7 / an^4 (a^3 minus a = a^2 same as n^7 minus n^4 = n^3)
=a^2n^3
19. -yz^5 / y^2z^3
= -z^2/y
20. -7x^5y^5z^4 / 21x^7y^5z^2 (divide -7 to 21 and minus xyz)
= -z / 3x^2
21. 9a^7b^5x^5 / 18a^5b^9c^3
=a^2c^2 / 2b^4
22. (n^5)^4
n ^5 x 4
=n^20
23. (z^3)^6
z ^3 x 6
=z^18
Answer:
Both the boats will closet together at 2:21:36 pm.
Step-by-step explanation:
Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).
Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),
Formula : d=v*t
at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)
the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)
Formula : D=
⇒ 
Now let 
∵ 
⇒ t= 450/1250
⇒ t= .36 hours
⇒ = 21 min 36 sec
Since F"(t)=0,
∴ This time gives us a minimum.
Thus, The two boats will closet together at 2:21:36 pm.
Answer:
A=2 (Wl+hl+hw)
Step-by-step explanation:
L=Length
W=Width
H=Height