6.) kp = k, tkz -2Nlm t 6 Nlm EN/m a:) L= 0-200 m X= 000 m T= 21 kp WINり -7 Toog m= 0·l00 kq ENh. Xi tYz = 30 cm X1 = 22.5 cm X2 = 7-5 cm %3D Ta 22:21 s aN/M X,= 3X2 %3D new equilibrium position is posiñion is 52.5 cm =0-523 m 30 cm t 22.5 cm = The length of each spning when the block is in Ik nen Post fion is 0 525 m
6.) kp = k, tkz -2Nlm t 6 Nlm EN/m a:) L= 0-200 m X= 000 m T= 21 kp WINり -7 Toog m= 0·l00 kq ENh. Xi tYz = 30 cm X1 = 22.5 cm X2 = 7-5 cm %3D Ta 22:21 s aN/M X,= 3X2 %3D new equilibrium position is posiñion is 52.5 cm =0-523 m 30 cm t 22.5 cm = The length of each spning when the block is in Ik nen Post fion is 0 525 m
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Please do help me in this physics questions please. I promise to give a good feedback and perfect score.. Please i badly need it in 1 hour from now. This is already an answered equations/problems.
All i need to know is to know if why is it kinetic energy etc. is indicated in here. And why the symbol in this particular parts is equals or plus and minus. Or a complete explanation on why this is my answer and where did i get it from.
PLease, i do hope you could help me. I promise to give a perfect score as a feedback.
Transcribed Image Text:2.
6.) kp= k, tkz - 2N/m + 6 Nlm
EN/m
%3D
9:) L 20om
X= 0100 m
と,- aN/m
T= 27
kp
WINO -ス
Toog
l00
m= 0·l00 tg
Ki Xo =kzXz
ZAHH X, = 6 Nm Yz
aN/n
X, = 3X2
Xi tXz = 30 cm
%3D
T= 22.21 s
2.5ג - ;X
X2 =7-5 cm
%3D
%3D
new equilibrium posiñion is
30 cm t 22.5 Cm - 52.5 cm = 0-523 m
i the length of cach cpring when the block is in
Its new Postfion is 0 525m
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