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alexandr1967 [171]
3 years ago
9

Suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of the object

varies inversely with its mass. a certain force acting upon an object with mass 3 kg produces an acceleration of 15 /ms2 . if the same force acts upon another object whose mass is 5 kg , what would this object's acceleration be?
Physics
1 answer:
amm18123 years ago
6 0
Solve for "x"
X = speed of the 5 kg 
3/15=5/x
cross multiply 
3x=75 
because (15)(5)=(3)(x)
divide 3 in both side
X=25 
or 
A=25 m/s^2 
Hope this helps

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I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

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Explanation:

Given data

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solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

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so that

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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

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