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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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Delicious77 [7]

Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

a(t) = -32 ft/s^2

v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

Find:-

a. Determine the position function s(t),

b. the velocity function v(t),

c. the acceleration function a(t).

d. What is the velocity of the ball at impact?

Solution:-

- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

                                  a(t) = d v(t) / dt

                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

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3 years ago
What is the answer to these questions?
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Answer:

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7 0
2 years ago
A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t
diamong [38]

Answer:

11.98 N

Explanation:

Normal force =   mg =  2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

       .602 = F / (2.03 * 9.81)   = 11.98  N

5 0
2 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = 6.38\times 10^6\ m

The gravitational force will balance the centripetal force

\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
What are the two principle advantages of telescopes over eyes?
Stells [14]

Answer:

i) Telescopes can be used to view far distant objects but the human eye can't view far distant objects.

ii) Telescopes uses two convex lenses producing a magnified image while the human eye only possesses one convex lens (image seen are smaller than that viewed under telescopes)

Explanation:

The telescopes can be used to view far distant objects due to their presence of two convex lenses. The two convex lenses are the objective lens (lens closer to object) and the eye piece lens (lens closer to eye). The object to be viewed forms an intermediate image first before the final image is seen using the eye piece lens.

The human eye only possess one convex lens and as such cannot view far ranged objects.

8 0
3 years ago
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