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3 years ago

Two stars that are in the same constellation:

Physics

1 answer:

Burka [1]3 years ago

7 0

Answer:

Option (3)

Explanation:

A constellation is usually defined as a collection of stars that appears in the sky forming a particular type of shape or pattern. They are located at a very large distance from one another. These constellations are often used in order to locate any astronomical objects in space.

These stars appearing in a constellation generally appears to be closer to one another in the night sky, but they are actually located at about many light years away from one another.

For example, in the Orion constellation, the stars appearing in the sky are about 25-30 light-years away from one another.

Thus, the correct answer is option (3).

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In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul

Bad White [126]

The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

Near field or fresnel. In this case the distance from the observer is small and we can see the shape of the object that creates the diffraction.

Far field or Fraunhoger. In this case the distance between the obstacle (leaves) and the person is great, here the information on the shape of things is lost and we have two observable forms. Lines for the case of slits and circles for the case of objects with a closed shape.

In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

The shape of the dots is circular because it is in the range of far-field diffraction.

Learn more about diffraction here: brainly.com/question/20140459

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2 years ago

A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of

kirill [66]

Answer:A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)

Explanation:

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3 years ago

Need the answer for question 5 :)

kykrilka [37]

Answer:

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3 years ago

What is the difference between: first moment of area and second moment of area?

zhannawk [14.2K]

To determine the centroid of the object first moment of area is used.

To predict the resistance of a shape to bending and deflection which are directly proportional, second moment of area is used.

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3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

1 year ago

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