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2 years ago

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An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 15m and a cross-section

al area 8.00cm^2. Take Young's modulus for steel to be Y= 2.0 x 10^11 Pa.
How much is the rod stretched (change in length of) when the ride is at rest? Assume that each car with two people seated in it has a total weight of 1900 N.

1 answer:

Oliga [24]2 years ago

7 0

Explanation:

Given that,

Length of each rod, L = 15 m

Area of cross section of the rod, $A=8\ cm^2=0.0008\ m^2$

Young's modulus for steel, $Y=2\times 10^{11}\ Pa$

Total weight of two people, F = 1900 N

We need to find the stretching in the rod when the ride is at rest. The formula of Young's modulus of a material is given by :

$Y=\dfrac{FL}{A\Delta L}$

Here, $\Delta L$ is the stretching in the rod

$\Delta L=\dfrac{FL}{YA}\\\\\Delta L=\dfrac{1900\times 15}{2\times 10^{11}\times 0.0008}\\\\\Delta L=1.78\times 10^{-4}\ m$

So, the rod will stretched to $1.78\times 10^{-4}\ m$

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