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find the point at {array}{l}x=y^{2} _{1}=0.5 . =1 {rad} / {sec

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find the point at $x = y^{2} μ_{1} = 0.5. w = 1 rad / sec a = 10 m / s^{2} . g = 10 m / s^{2} . m = 0.001 kg$ which the bead fy.

Name: Video solution 1: find the point at x=y2μ1=0.5.w=1rad/seca=10 m/s2.g=10 m/s2.m=0.001 kg which the bead fy.
Uploaded: 2023-02-06T15:03:49.000Z
Duration: 6 min
Description: Video solution 1: find the point at x=y2μ1=0.5.w=1rad/seca=10 m/s2.g=10 m/s2.m=0.001 kg which the bead fy.

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Updated on: Feb 6, 2023

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find the point at

x = y^{2} μ_{1} = 0.5. w = 1 rad / sec a = 10 m / s^{2} . g = 10 m / s^{2} . m = 0.001 kg

which the bead fy.

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Question Text	find the point at $x = y^{2} μ_{1} = 0.5. w = 1 rad / sec a = 10 m / s^{2} . g = 10 m / s^{2} . m = 0.001 kg$ which the bead fy.
Updated On	Feb 6, 2023
Topic	Thermodynamics
Subject	Physics
Class	Class 11
Answer Type	Video solution: 1
Upvotes	82
Avg. Video Duration	6 min

find the point at $x = y^{2} μ_{1} = 0.5. w = 1 rad / sec a = 10 m / s^{2} . g = 10 m / s^{2} . m = 0.001 kg$ which the bead fy.

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find the point at x=y2μ1​=0.5.w=1rad/seca=10 m/s2.g=10 m/s2.m=0.001 kg​ which the bead fy.

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Practice more questions on Thermodynamics

Students who ask this question also asked

find the point at $x = y^{2} μ_{1} = 0.5. w = 1 rad / sec a = 10 m / s^{2} . g = 10 m / s^{2} . m = 0.001 kg$ which the bead fy.