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5. A 2 kg mass is attached to a massless spring of force constant k25 N/m. The spring is stretched 0,40 m from its equilibriu
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0. Mazky, kz 25 Mlon 4x = 0,40m Energy E= 1/2k (Ax)² as E = * 25 *(0.4072 frequery. It a les HO E = (n+16) tw - 2 = (n+1, 1.Eos ħw Eos 1.06 Xiom X 2at 84 to = 1.06X164 x 242.14 X3.535 E = 2.3531 x 1033 Joule

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