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3 years ago

What is the sidereal period used in Kepler’s third law?

1 answer:

3 0

His law exaplins/shows that the average distance of a planet from the Sun cubed is directly proportional to the orbital period squared.

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Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Photons: particles that constitute light. </em>

3 0

3 years ago

Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I

mezya [45]

The problem is basically asking us to find a way to find the sound intensity I, in terms dependent on the sound level and the reference intensity $I_0$ .For this purpose we can start from the unit used in the scale logarithmic decibel, that is

$\beta = 10log_{10}\frac{I}{I_0}$

Where

I = Acoustic intensity on the linear scale

$I_0 =$ Hearing threshold

Using the logarithmic properties of the exponents the above expression can be described as:

$(\frac{I}{I_0})^{10} = 10^{\beta}$

$I = I_0 10^{\frac{\beta}{10}} \righarrow$ that is the expression or technique to find the intensity of sound.

8 0

3 years ago

A 2.75 kg block is pulled across a flat, frictionless floor with a 5.11 N force directed 53.8° above horizontal. What is the tot

Softa [21]

Answer:

F = 3.01 N

Explanation:

Given that,

Mass of a block, m = 2.75 kg

Force applied to the block, F = 5.11 N

It is directed 53.8° above horizontal.

We need to find the total force acting on the block. The force acting on it is given by :

$F=F\cos\theta\\\\F=5.11\times \cos53.8\\F=3.01\ N$

So, 3.01 N of force is acting on the block.

4 0

3 years ago

What is the equilibrium constant of pure water at 25C

Tcecarenko [31]

Answer:

1.0 x 10-14.

Explanation:

We then replace the term on the right side of this equation with a constant known as the water dissociation equilibrium constant, Kw. In pure water, at 25C, the [H3O+] and [OH-] ion concentrations are 1.0 x 10-7 M. The value of Kw at 25C is therefore

6 0

4 years ago

A large aquarium of height 6 m is filled with fresh water to a depth of D = 1.50 m. One wall of the aquarium consists of thick p

Vesnalui [34]

Answer

given,

height of aquarium = 6 m

Depth of fresh water = D = 1.50 m

horizontal length of the aquarium(w) = 8.40 m

total force increased when liquid is filled to depth = 4.30 m

g = 9.81 m/s²

ρ = 998 Kg/m³