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3 years ago

Why do the planets appear in different locations in the night sky while the pattern of stars in a constellation stays the same?

2 answers:

8 0

Planets are consantly moving and are relitivly nearby so their movment has a big impact on our veiw o them but stars are far away so it take a huge amount of movement to change our view of them

Goshia [24]3 years ago

7 0

I'm sure you've noticed that an airplane high in the sky, far away
from you, looks like it's moving very slowly. At the same time,
somebody passing you on a skateboard whizzes past you at
high speed. The farther away something is from you, the slower
it appears to move.

The nearest star outside the solar system is almost 32 thousand times
as far away from us as the farthest visible planet (Saturn) is, and all of the
other stars are farther than that.

That's why you have to wait a few thousand years before you notice
that the shape of a constellation has changed.

To put it a slightly different way . . . Everything is in motion. The motion is
more noticeable for nearby things, and less noticeable for farther-away things.
Objects within our solar system are the only ones near enough so that a human
lifetime is a long enough period in which to notice the change in their position.
Even Pluto moves less then 1.5° against the 'background' stars in a whole year.

This all makes me feel small. How about you ?

You might be interested in

Identify the areas on the image where the force of repulsion is the least.

aalyn [17]

Answer:

The central blue square in between the lower pair of magnet has the least force of repulsion.

Explanation:

We can explain this using the dual nature of magnets.

Each magnet must have two poles namely:

-North pole

-South pole

We assume that the magnetic lines of forces enters from south pole and leaves from the north pole.

When brought together, like poles repel each other while opposite poles attract each other.

In the picture, the lower two magnets have opposite poles facing each other, hence the force of repulsion is minimum there and the force of attraction is maximum.

4 0

3 years ago

Read 2 more answers

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

2 years ago

Read 2 more answers

2. What will be the extension of this spring if the load is a) 4N and b) 75 g?

Furkat [3]

Answer:

Explanation:

just add

7 0

3 years ago

PLZZZZZ HELP ASAP I WILL MARK BRAINIEST FOR WHOEVER HAS THE BEST ANSWER!!!!!! THIS IS MY THIRD TIME PUTING THIS QUESTION UP I DO

ddd [48]

Answer:

Well first for criteria think what would the rover need in order to sustain itself on Venus. And for constraints think of anything that could possibly affect the rover( ex: gasses, active volcanoes)

Explanation:

Criteria: Make the rover self sustainable, and allow the rover to have a mission on Venus( ex: collect rock samples)

Constraints, as I mentioned above gasses, and active volcanoes.

I hope this helps! :)

8 0

3 years ago

Can I get some help? I REALLY need it!

IceJOKER [234]

B. Because they are used to see through objects

8 0

3 years ago