परी भ्रमण त्रिज्या से आपका क्या तात्पर्य है 5.00 सीएम वाली ध्रुव की परिभ्रमण त्रिज्या केंद्र से गुजरने वाली एंव उसके तल के लंबवत अक्ष के परित: परिकलित कीजिए
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Given 5. Calculate the radius of gyration of a hoop with a radius of 5.00 cm passing through the center and normal to the axis of its plane.
- We need to find the radius of gyration of a hoop of radius 5 cm rotating about an axis passing through its centre and normal to its plane
- So moment of inertia about a tangent of the sphere will be
- 2 MR^2 / 5 + MR^2
- = 7 MR^2 / 5
- Or MK^2 = 7 MR^2 / 5
- Or K^2 = 7 R^2 / 5
- Or K ^2 = 7 x 5^2 / 5
- Or K^2 = 35
- Or K = √35 cm
Reference link will be
https://brainly.in/question/11427722
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