Please solve this in step by step process and box the 3 ANSWERS. Please draw an illustration also Two objects are connected by a lightstring that passes over a frictionlesspulley as shown. Assume the inclineis frictionless and take m1=2.00 kg,m2=6.00 kg, and e=55° . Draw free-body diagrams of both objects. Find(a) the magnitude of the accelerationof the objects, (b) the tension in thestring, and (c) the speed of eachobject 2.00 s after it is released fromrest.Choose 3 answers--- Xm2gsinom2gcos 6mg Choose 3 answersm2M )--- xm2gsin@m2gcos 0vf = 3.14 m/sa =1.57 m/s^2OT = 26.7 Nvf = 7.14 m/s%3D= 3.57 m/s^2a =T = 16.7 N%3D

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Answered: Two objects are connected by a light…

Two objects are connected by a light string that passes over a frictionless pulley as shown. Assume the incline is frictionless and take m1=2.00 kg,

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter5: Newton's Laws Of Motion

Section: Chapter Questions

Problem 23PQ: The x and y coordinates of a 4.00-kg particle moving in the xy plane under the influence of a net...

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Special Relativity

The theory of special relativity discusses how space and time are related to objects traveling along a straight line at a constant speed. Simply stated, as an object reaches the speed of light, its mass becomes infinite and it is unable to move faster than light. This interstellar speed limit has been the topic of much debate in astronomy, and also in science fiction when people consider how to fly over large distances.

Question

Please solve this in step by step process and box the 3 ANSWERS. Please draw an illustration also

Two objects are connected by a light
string that passes over a frictionless
pulley as shown. Assume the incline
is frictionless and take m1=2.00 kg,
m2=6.00 kg, and e=55° . Draw free-
body diagrams of both objects. Find
(a) the magnitude of the acceleration
of the objects, (b) the tension in the
string, and (c) the speed of each
object 2.00 s after it is released from
rest.
Choose 3 answers
--- X
m2gsino
m2gcos 6
mg

Choose 3 answers
m2
M )--- x
m2gsin@
m2gcos 0
vf = 3.14 m/s
a =
1.57 m/s^2
OT = 26.7 N
vf = 7.14 m/s
%3D
= 3.57 m/s^2
a =
T = 16.7 N
%3D

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