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andreev551 [17]
2 years ago
13

Help!!!!!!!!!!!!!!!!!​

Physics
2 answers:
Illusion [34]2 years ago
6 0

Answer:

(4) Elastomers

Explanation:

Substances which can be elastically stretched to large value of strain are called elastomers.

Fudgin [204]2 years ago
5 0

Answer:

<h3>Substances which can be easily stretched to large value of strain are called</h3><h2><em>Elastomers</em></h2>

<h3><em>Some</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>Examples are the tissue of the aorta, rubber, etc.</em></h3>
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People, especially adults, are inactive because of "passive" modes of transportation.
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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
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Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

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Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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