Asked by melissa
An air puck of mass m1 = 0.45 kg is tied to a string and allowed to revolve in a circle of radius R = 1.0 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass m2 = 1.10 kg is attached to it. The suspended mass remains in equilibrium while the puck on the tables revolves.
A) What is the tension T in the string?
B) What is the magnitude of the horizontal force acting on the air puck?
C) What is the speed of the puck
A) What is the tension T in the string?
B) What is the magnitude of the horizontal force acting on the air puck?
C) What is the speed of the puck
Answers
Answered by
Anonymous
Teresa's class?
Answered by
Brianna
a)What is the tension T in the string?
in this case, tension = mass2 X gravity
so, tension = 1.10 kg X 9.8 m/s^2
tension = 10.78
For letter b, I am afraid I have no clue :(
c)What is the speed of the puck?
F = (m(v^2))/r
the tension is the force in this case, so:
tension = (m1(v^2))/r
v^2 = (tension X r)/m1
v^2 = (10.78 N X 1.0 m)/ .45 kg
v^2 = 23.95555556 m/s
Take the square root of both sides
v = 4.894441291 m/s
in this case, tension = mass2 X gravity
so, tension = 1.10 kg X 9.8 m/s^2
tension = 10.78
For letter b, I am afraid I have no clue :(
c)What is the speed of the puck?
F = (m(v^2))/r
the tension is the force in this case, so:
tension = (m1(v^2))/r
v^2 = (tension X r)/m1
v^2 = (10.78 N X 1.0 m)/ .45 kg
v^2 = 23.95555556 m/s
Take the square root of both sides
v = 4.894441291 m/s
Answered by
Jake Conner
part a is the same as part b ;)
Answered by
Lisa
For B, the magnitude is simply the absolute value of what you got in part A.
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