Both synchronised formulas are revealed listed below. |
24 |
m/s = 2.0 – v1‘ + v2‘ ( 1 ) |
288 m2/ s2 = 2.0 – v1‘2 + v2‘2( 2 ) |
Solving
|
( 1) for v2‘ returns: |
v2‘ = 24 m/s – 2.0 – v1‘ |
Replacing the above expression for v 2’right into(2)returns: 288 m 2/ s 2= 2.0 – v 1 ‘2+(24 m/s-2.0 – v 1’
|
)2Increasing the above expression as well as streamlining returns: 288 m 2/ s 2 = 2.0 – v1‘2 |
+ |
(576 m 2/ s 2 -96 m/s – v 1’+4.0 v 1′ 2)288 288 m 2/ s 2=2.0 – v
|
1’2+576 m 2/ s 2-96 m/s – v 1′ +4.0 v 1′2 v 1’2– 16 m/s – v1‘ + 48= 0Making use of theTI-83 Solver provides v |
1 ‘ =4.0 m/s as well as |
replacing this worth right into (1)offers v2‘ =16 m/s.
Related
No Result
View All Result
Copyright © 2022 All Rights Reserved | Powered by Mmsphyschem
|