Two masses of mass 3kg and 4kg are attached to the end of the string passed over a pulley fixed at the top .the tension and accleration are
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m1 = 2kg
m2= 3kg
Now,
T= (2m1 × m2 / m1 × m2 )g
= (2 ×2 ×3/ 2+3 )g
= 12/5 g
a = (m2 - m1 / m1+ m2) g
= (3-2/3+2)g
= g/5
Answered by
1
Answer:
Explanation:m1 = 2kg
m2= 3kg
Now,
T= (2m1 × m2 / m1 × m2 )g
= (2 ×2 ×3/ 2+3 )g
= 12/5 g
a = (m2 - m1 / m1+ m2) g
= (3-2/3+2)g
= g/5
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