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jolli1 [7]
3 years ago
7

In a horse cart system when would there be the smallest amount of acceleration

Physics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

well... when the horse stops/rests, or if it is blocked by a surface or anything of solid background.

Explanation:

If it is going up a hill or slope and it just starts to move that would also be considered the smallest amount of acceleration this can go for many things when it just starts to move. but I would go for when it rests amounting to your fitting of the question.

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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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