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Andru [333]
3 years ago
7

Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

ck telescope?
Physics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

4 times greater

Explanation:

<u>Step 1:</u> Calculate light-collecting area of a  20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² =\frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 20-m

A_{1} = \frac{\pi d^{2}}{4}

A_{1} = \frac{\pi (20^{2})}{4}

A₁ = 314.2 m²

<u>Step 2:</u> Calculate light-collecting area of a  10-meter Keck telescope (A₂)

A_{2} = \frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 10-m

A_{2} = \frac{\pi (10^{2})}{4}

A₂ = 78.55 m²

<u>Step 3</u>: divide A₁ by A₂  

= \frac{314.2 m^2}{78.55 m^2}

= 4

Therefor,  the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

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Lynna [10]

Answer:

Normal Force is usually perpendicular to the movement and static friction usually means that there is no movement.

Explanation:

The work donde by any force on an object is equal to the displacement of the object multiplied by the component of the force that is in the direction of the displacement.

Normal force is usually perpendicular to the movement, so there is no component in the direction of the displacement. This is why it is zero in most circumstances.

<em>Static</em> friction on the other hand, usually means that there is no movement at all (it's static). It means that there is no displacement between the object and ground (in most cases). If there is no displacement, there is no work.

4 0
3 years ago
La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0
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Schach [20]

Answer:

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Explanation:

For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane

Y axis

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       N = mg cos θ

X axis

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       mg sin θ = m a

        a = g sin θ

let's calculate

        a = 9.8 cos 40

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