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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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Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variabl
Shtirlitz [24]

Answer:

D)     D = \frac{5}{4}  - \frac{3}{4} \ C, E)  (C, D) = ( \frac{17}{7}, \ \frac{-4}{7}

Explanation:

Part D) two expressions are indicated

          3C + 4D = 5

          2C +5 D = 2

let's simplify each expression

         3C + 4D = 5

         4D = 5 - 3C

we divide by 4

            D = \frac{5}{4}  - \frac{3}{4} \ C

The other expression

       2C +5 D = 2

       2C = 2 - 5D

        C = 1 -  \frac{5}{2} \ D

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

          D =  \frac{5}{4}  - \frac{3}{4} \ C

we substitute in the other equation

            2C +5 D = 2

             2C +5 (5/4 - ¾ C) = 2

we solve

            C (2 - 15/4) + 25/4 = 2

             -7 / 4 C = 2 - 25/4

             -7 / 4 C = -17/4

              7C = 17

               C = \frac{17}{7}

now we calculate D

               D = \frac{5}{4} - \frac{3}{4} \ \frac{17}{7}

               D = 5/4 - 51/28

               D =\frac{35-51}{28}

               D = - 16/28

               D = - \frac{4}{7}

the result is (C, D) = ( \frac{17}{7}, \ \frac{-4}{7} )

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3 years ago
Does magnets exert a force. explain​
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7 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

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where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

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\theta=\frac{3.7m}{0.162m}=22.83rad

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Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

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3 years ago
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