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kykrilka [37]
3 years ago
12

Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing

through its center.
Physics
1 answer:
Juliette [100K]3 years ago
3 0

Answer: MR²

is the the moment of inertia  of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements  are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

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Explanation:

Given that,

Initial speed of the billiard ball 1, u = 30i cm/s

Initial speed of another billiard ball 2, u' = 40j cm/s

After the collision,

Final speed of first ball, v = 50 cm/s

Final speed of second ball, v' = 0 (as it stops)

Let us consider that both balls have same mass i.e. m

Initial kinetic energy of the system is :

K_i=\dfrac{1}{2}mu^2+\dfrac{1}{2}mu'^2\\\\K_i=\dfrac{1}{2}m(u^2+u'^2)\\\\K_i=\dfrac{1}{2}m((30)^2+(40)^2)\\\\K_i=1250m\ J

Final kinetic energy of the system is :

K_f=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv'^2\\\\K_f=\dfrac{1}{2}m(v^2+v'^2)\\\\K_f=\dfrac{1}{2}m((50)^2+(0)^2)\\\\K_f=1250m\ J

The change in kinetic energy of the system is equal to the difference of final and initial kinetic energy as :

\Delta K=K_f-K_i\\\\\Delta K=1250m-1250m\\\\\Delta K=0

So, the change in kinetic energy of the system as a result of the collision is equal to 0.    

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4 years ago
8.How long is a day? A year?
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Answer:

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A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase
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Answer:

Y_A=277.128 \angle 30v

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From the question we are told that

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Generally voltage drop across phase B

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Y_B=277.128 \angle (-150)v

Generally voltage drop across phase C

Y_C=277.128 \angle (-30+120)

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