please check my work if this is correct thank you! 3:EF= maa= ZEMa= my g-m₂9-X₁ + 7/M₁ + m₂q= m, g-m₂ 1m₁ + m₂a=9/ M₁-M₂mitme9 = (2.1-1.9) 9.82-1 +19-+= √2² = √(2)(0.2m)0.49 m/s²15SM2mi-0.49 m/s²/0-904-8NO.:DATE: 3. Two masses in an Atwood machine are 1.9 and 2.1 kg, the vertical distance ofthe heavier body being 20 cm above the lighter one. After what time would thelighter body be above the heavier one by the same vertical distance? Neglect themass of the pulley and the cordTmimg

3. Two masses in an Atwood machine are 1.9 and 2.1 kg, the vertic the heavier body being 20 cm above the lighter one. After what tin lighter body be above the heavier one by the same vertical distanc mass of the pulley and the cord

3. Two masses in an Atwood machine are 1.9 and 2.1 kg, the vertic the heavier body being 20 cm above the lighter one. After what tin lighter body be above the heavier one by the same vertical distanc mass of the pulley and the cord

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter4: Newton's Laws Of Motion

Section: Chapter Questions

Problem 40P: The leg and cast in Figure P4.40 weigh 220 N (w1). Determine the weight w2 and the angle needed so...

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Question

please check my work if this is correct thank you!

3:
EF= ma
a= ZE
M
a= my g-m₂9-X₁ + 7/
M₁ + m₂
q= m, g-m₂ 1
m₁ + m₂
a=
9/ M₁-M₂
mitme
9 = (2.1-1.9) 9.8
2-1 +19-
+= √2² = √(2)(0.2m)
0.49 m/s²
15
S
M2
mi
-0.49 m/s²
/0-904-8
NO.:
DATE:

3. Two masses in an Atwood machine are 1.9 and 2.1 kg, the vertical distance of
the heavier body being 20 cm above the lighter one. After what time would the
lighter body be above the heavier one by the same vertical distance? Neglect the
mass of the pulley and the cord
T
mi
mg

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