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Karo-lina-s [1.5K]
2 years ago
12

Pls answer ASAP pls bc I’m tryna get my grade up please

Physics
1 answer:
Sidana [21]2 years ago
6 0

Answer:

The right answer for this question is 85%.

(I had the same question.)

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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