Asked by savannah
Two objects with masses of 3.00 kg and 7.00 kg are connected by a light string that passes over a frictionless pulley
(a) Determine the tension in the string.
.N
(b) Determine the magnitude of the acceleration of each mass. .m/s2
(c) Determine the distance each mass will move in the first second of motion if both masses start from rest.
m
(a) Determine the tension in the string.
.N
(b) Determine the magnitude of the acceleration of each mass. .m/s2
(c) Determine the distance each mass will move in the first second of motion if both masses start from rest.
m
Answers
Answered by
Bri
A.
m1a1=T-m1g and m2a2=T-m2g are your two equations, substitute and add and you get this -->
T= (2xm1xm2)/(m1+m2) all multiplied by 9.8 (g).
T= (2(3.0)(7.0))/(3+7) = 4.2(9.8) = 41.16 N
B.
Then just plug it in to one of your original equations.
m1a1 = T-m1g
(3.0)a = (41.16N)-(3.0)(9.8)
a= 3.92 m/s squared.
C. First use equation Vf = Vi + at
Vf = (0) + (3.92)(1)
Vf= 3.92 m/s
Then plug that into V(squared) = Vi(squared) + 2aX
(3.92)squared = 0(squared) + 2 (3.92)X
X= 1.96m
m1a1=T-m1g and m2a2=T-m2g are your two equations, substitute and add and you get this -->
T= (2xm1xm2)/(m1+m2) all multiplied by 9.8 (g).
T= (2(3.0)(7.0))/(3+7) = 4.2(9.8) = 41.16 N
B.
Then just plug it in to one of your original equations.
m1a1 = T-m1g
(3.0)a = (41.16N)-(3.0)(9.8)
a= 3.92 m/s squared.
C. First use equation Vf = Vi + at
Vf = (0) + (3.92)(1)
Vf= 3.92 m/s
Then plug that into V(squared) = Vi(squared) + 2aX
(3.92)squared = 0(squared) + 2 (3.92)X
X= 1.96m
Answered by
Mohammed
This answer is complete trash
Answered by
Josiah
Perfect, thanks!!
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