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3 years ago

14

un esquiador parte del reposo y se desliza pendiente abajo recorriendo 9m en 3s, con una aceleración constante calcular acelerac

ión y el tiempo que tardará en adquirir la velocidad de 24m/s con la misma aceleración

1 answer:

Troyanec [42]3 years ago

7 0

De verdad no estoy seguro pero creo q es 72 segundos porque 9*24 es 216 y cuando lo divides por 3 es 72 segundos

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2 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

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Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

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$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

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