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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

The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov

victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

$T_{1} + V_{1} = T_{2} + V_{2}$

$h_{ab} = h_{bc} = 0.18m$

potential energy at position 1,

$V1 = m_{ab} gh_{ab} + m_{bc} gh_{bc}$

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

$T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} + \frac{1}{2} m_{bc} v^{2}_{G} + \frac{1}{2} lw^{2}_{bc}$

$l_{ab} =\frac{m_{ab} l^{2}_{ab} }{3}$

= 1/3 *4 * (0.36)²

=0.10368kg m²

$l =\frac{m_{bc} l^{2}_{bc} }{12}$

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

#SPJ4

5 0

1 year ago

An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible

harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

3 0

3 years ago

What process is mostly responsible for the blue appearance of the sky

malfutka [58]

Answer:reflection by dust particles in air

7 0

3 years ago

Plz help me! <br><br> How do electromagnetic fields work?

MrMuchimi

Answer:

Electromagnetic field, a property of space caused by the motion of an electric charge. A stationary charge will produce only an electric field in the surrounding space. If the charge is moving, a magnetic field is also produced. An electric field can be produced also by a changing magnetic field.

5 0

3 years ago

Estimate the total number of bacteria and other prokaryotes in the biosphere of the earth. (assume the bacteria are found to a d

Fofino [41]

Since the Earth is almost spherical in shape, we are actually to find first the volume of the spherical segment at a depth of 1,000 m. The radius of the Earth is 6,371,000 meters. The volume of a spherical segment is:

V = 1/3*πh²(3r - h)
Substituting the values and making sure the units is in mm,
V = 1/3*π(1000 m * 1000 mm/1 m)²[3(6,371,000 m * 1000 mm/1 m) - (1000 m * 1000 mm/1 m)]
V = 2×10²² mm³

Thus, the total amount of bacteria is:

2×10²² mm³ * 100 bacteria/1 mm³ = 2×10²⁴ bacteria

7 0

3 years ago