Asked by kwaku
A 0.2 kg mass hangs at the end of a wire. what is the tension in the wire if the ball is whirled around in a horizontal circle with tangential velocity of 5 m/sec
Answers
Answered by
drwls
Let the tension be T and the angle of the string to the horizontal be A.
Write equations for horizontal and vertical motion.
T sin A = M g
T cos A = M V^2/r
Square both sides and add the two equations to get T
T^2(sin^2A + cos^2A) = T
= M^2(g + V^2/r)
Solve for T.
You need to know the length of the wire (r) to solve this problem. They should have told you what it is.
Write equations for horizontal and vertical motion.
T sin A = M g
T cos A = M V^2/r
Square both sides and add the two equations to get T
T^2(sin^2A + cos^2A) = T
= M^2(g + V^2/r)
Solve for T.
You need to know the length of the wire (r) to solve this problem. They should have told you what it is.
Answered by
Ann
A person places the speakers 4.0 m apart and connects a signal generator to both speakers that produces a single and consistent tone. (constant wavelength and frequency) He then walks to a point that is 2.0 m from one speaker and 2.3 m from the other. At that point he notices a quiet “spot”. If the speed of the sound in the room is known to be 350 m/s, calculate the possible frequencies being played by the speakers.
My Answer:
PD=2.3m-2m
PD=.3m
PD=(n-.5) λ, but λ=v/f
PD=(n-.5)(v/f)
.3=(n-.5)(350/f)
f=(n-.5)(350/.3)
f=(n-.5)(3500/3)
f=(3500n/3)-(3500/6)
f=(3500n/3)-(3500/6), where n is any real integer.
Is this right? Thanks for your help.
My Answer:
PD=2.3m-2m
PD=.3m
PD=(n-.5) λ, but λ=v/f
PD=(n-.5)(v/f)
.3=(n-.5)(350/f)
f=(n-.5)(350/.3)
f=(n-.5)(3500/3)
f=(3500n/3)-(3500/6)
f=(3500n/3)-(3500/6), where n is any real integer.
Is this right? Thanks for your help.
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