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3 years ago

6

un resorte de 10cm de longitud recibe una magnitud de fuerza que lo estira hasta medir 15cm ¿cual es la magnitud de la tension u

nitaria o deformacion lineal? Si me explican la fórmula estaría genial!

1 answer:

Ber [7]3 years ago

8 0

Answer: 0.5

Explanation:

The modulus of elasticity (called <em>"alargamiento unitario"</em> in spanish) $\epsilon$ of a spring is given by the following formula:

$\epsilon=\frac{\Delta L}{L}$

Where:

$L=10 cm$ is the original length of the spring

$\Delta L=L_{f}-L$ is the elongation of the spring, being $L_{f}=15 cm$ the length of the spring after a force is applied to it.

$\epsilon=\frac{L_{f}-L}{L}=\frac{15 cm - 10 cm}{10 cm}$

Then:

$\epsilon=0.5$

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How to Answer questions 12 and 13

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Explanation:

12)

The weight of an object is equal to the force of gravity acting on the object. Near the Earth's surface, it can be calculated as

$W=mg$

where

W is the weight

m is the mass of the object

g is the acceleration of gravity

In this problem, we know that the mass of the plane is

m = 75 tonnes

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13)

We can solve this question by applying Newton's second law along the vertical direction of motion. In fact, the net force acting along the vertical direction must be equal to the product between the mass of the plane and the vertical acceleration:

$\sum F_y = ma_y$

where

$\sum F_y$ is the net force in the y-direction

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$a_y$ is the acceleration in the y-direction

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So the vertical net force is

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