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EleoNora [17]
3 years ago
9

Frente a una lente convergente delgada se coloca un objeto a una distancia de 50 cm. La imagen de este objeto aparece del otro l

ado a 80 cm de la lente. ¿Cuál es la distancia focal de la lente?​
Physics
1 answer:
katovenus [111]3 years ago
4 0

Answer:

La distancia focal del objetivo es de aproximadamente 30,77 cm

Explanation:

Los parámetros dados son;

La distancia del objeto desde la lente, u = 50 cm

La distancia de la imagen desde la lente, v = 80 cm

La fórmula del espejo se puede escribir de la siguiente manera;

\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}

Dónde;

f = La distancia focal de la lente

v = La distancia de la imagen desde la lente

u = La distancia del objeto desde la lente

Sustituyendo los valores da;

\dfrac{1}{f} = \dfrac{1}{80} + \dfrac{1}{50} = \dfrac{80 + 50}{4000} = \dfrac{13}{400}

\therefore  f = \dfrac{400}{13}  = 30\dfrac{10}{13} \ cm

La distancia focal de la lente, f = 30¹⁰/₁₃ cm ≈ 30,77 cm.

La distancia focal del objetivo ≈ 30,77 cm.

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Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

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Explanation:

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So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

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