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Anika [276]
3 years ago
9

Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an ax

is perpendicular to the plane of the square and passing through its center

Physics
1 answer:
deff fn [24]3 years ago
7 0

Answer:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m  

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I =  3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

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What is the impulse of a 3kg object accelerating from rest to 12m/s?
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Why does a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s?
andrew11 [14]

A dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.

s = vt - 1 / 2 at²

s = Displacement

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The equation used to solve the given problem is an equation of motion. In a free fall motion, usually air resistance is not considered for easier calculation. If air resistance is considered acceleration cannot be constant throughout the entire motion.

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brainly.com/question/5955789

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