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allsm [11]
2 years ago
8

A hole of radius r has been drilled in a circular, flat plate of radius R. The center of the hole is a distance d from the cente

r of the circle. You can assume that the mass of the solid disk before the hole is removed was M. Find the moment of inertia for the rotation about an axis through the center of the circle, perpendi
Physics
1 answer:
Setler [38]2 years ago
4 0

Answer:

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

Explanation:

From the question, we have to consider the hole as an object of negative mass. Thus, the total inertia of the object will be;

I_total = I_plate - I_hole

Moment of inertia of a Disk is ½MR²

We are told it rotates around it's axis. Hence, we will use parallel axis theorem and we have;

I_plate = ½M_plate•R²

I_hole = ½M_hole•r² + M_hole•d²

I_hole = ½M_hole(r² + 2d²)

Thus;

I_total = ½M_plate•R² - ½M_hole(r² + 2d²)

Now,since the plate is uniform, then;

M_hole/M_plate = A_hole/M_plate

A_hole = πr²

A_plate = πR²

Thus;

M_hole/M_plate = πr²/πR²

M_hole/M_plate = r²/R²

M_hole = M_plate(r²/R²)

Let's put M_plate(r²/R²) for M_hole in the I_total equation to get;

I_total = ½M_plate•R² - (½M_plate(r²/R²)(r² + 2d²))

Factorizing, we have;

I_total = ½M_plate[R² - ((r²/R²)(r² + 2d²))]

We are told that mass of the solid disk before the hole is removed was M.

Thus;

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

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The resistance expected of the heater is 50.1 ohms.

<h3>What is resistance?</h3>

Resistance can be defined as the opposition to the flow of electric current in an electric circuit. The S.I unit of resistance is Ohms (Ω).

To calculate the resistance of the heater, we use the formula below.

<h3>Formula:</h3>
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Where:

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From the question,

Given:

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Substitute these values into equation 1

  • R = (480²)/4600
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Hence, the resistance expected of the heater is 50.1 ohms.

Learn more about resistance here: brainly.com/question/17563681

3 0
2 years ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
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To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

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F_p = 9*10^{22}N

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F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

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Replacing with the previous force,

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Replacing our values

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F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

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