Answer:
L= 1.468 m
Explanation:
The moment of inertia of the rod about its center is (1/12)m_RL^2
The moment of inertia of each of the two bodies about the described axes is m_B(L/2)2
Hence, the moment of inertia of three body system is (1/12)m_RL^2+ 2×m_B(L/2)^2 which is given to be equal to I_T
=> L^2[(1/12)m_R+m_B/2] = I_T
=> L2 = IT/(mR/12+ mB/2)
=> L = sqrt( 12I_T/(m_R+6m_B))
now putting the value of m_R= 3.47 kg
m_B= 0.263 kg
I_T = 0.97 kg/m^2
![L= \sqrt{\frac{12I_T}{m_R+6m_B} }](https://tex.z-dn.net/?f=L%3D%20%5Csqrt%7B%5Cfrac%7B12I_T%7D%7Bm_R%2B6m_B%7D%20%7D)
![L= \sqrt{\frac{12\times0.907}{3.47+6\times0.263} }](https://tex.z-dn.net/?f=L%3D%20%5Csqrt%7B%5Cfrac%7B12%5Ctimes0.907%7D%7B3.47%2B6%5Ctimes0.263%7D%20%7D)
L= 1.468 m is the length of the rod be so that the moment of inertia of the three-body system
Longggggg duhhhhhh !!!!!!!!
![\huge{ \underline{ \boxed{ \bf{ \blue{Solution:}}}}}](https://tex.z-dn.net/?f=%20%5Chuge%7B%20%5Cunderline%7B%20%5Cboxed%7B%20%5Cbf%7B%20%5Cblue%7BSolution%3A%7D%7D%7D%7D%7D)
<h3>
<u>Provided</u><u>:</u><u>-</u></h3>
- Initial velocity = 15 m/s
- Final velocity = 10 m/s
- Time taken = 2 s
<h3><u>To FinD:-</u></h3>
- Accleration of the particle....?
<h3>
<u>How</u><u> </u><u>to</u><u> </u><u>solve</u><u>?</u></h3>
We will solve the above Question by using equations of motion that are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
Here,
- v = Final velocity
- u = Initial velocity
- a = acceleration
- t = time taken
- s = distance travelled
<h3>
<u>Work</u><u> </u><u>out</u><u>:</u></h3>
By using first equation of motion,
⇛ v = u + at
⇛ 10 = 15 + a(2)
⇛ -5 = 2a
Flipping it,
⇛ 2a = -5
⇛ a = -2.5 m/s² [ANSWER]
❍ Acclearation is negative because final velocity is less than Initial velocity.
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Stems and leaves is the answer. Jessica you are in my class right ?
Where are the answer choices. I wont be able to help without the answer choices or a picture