If Earth's axis was "straight up and down" instead of tilted, then ...
<span>-- There would be no seasons.
-- The climate at any one place would be the same all year around.
-- The days would be the same length, everywhere,
and all year around.
-- So would the nights.
-- The sun would be up a little more than 12 hours every day.
It would be down a little less than 12 hours every day.
-- At the middle of the day, the sun would be at the same height
in the sky all year around, not higher in some months and lower
in others.
-- The equator would be the only place on Earth where the sun
could ever be directly over your head.
-- If you were at the north pole or the south pole, the sun would be
down on the horizon, and it would just go around and around you
every day. It would never rise or set, and it would never get any
higher or lower.
</span>
Answer:
The galaxy is moving away from the observer
Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.
So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.
Answer:
Momentum P is 840000kgm/s or 8.4 × 10^6
Explanation:
Data :
Mass = 21000 kg
Velocity = 400 m/s
So momentum is given as
P = mv
P = 21000×400
P = 8400000 kgm/s
P = 8.4 × 10^6
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Before Pluto was discovered, it was predicted. Astronomers had observed that massive objects can affect the orbits of its neighbors, and, after seeing deviations in the orbits of Uranus and Neptune, assumed something substantial existed beyond their orbits.
When Pluto was spotted, it was thought to be the predicted object and was identified as a ninth planet.
A few decades later, astronomers started discovering more and more objects around other stars and didn’t know whether to call them planets or not. There appeared to be a need to define what a planet means, and that led to what some people consider Pluto’s demotion to a dwarf planet.
The International Astronomical Union decided that full-sized planets must orbit the sun, have a round shape, and have cleared their orbits of other objects. Pluto fulfills the first two criteria, but not the third.
It still goes around the sun, it’s round enough, it’s got moons, and behaves like a planet, but the idea is that Pluto did not form the same way as the rest of the planets. Pluto’s orbit is both eccentric and inclined more than the rest of the planets by about 17 degrees. That’s suggests something is different about this object.
This debate about whether to call it a planet or not is silly, because it doesn’t matter to Pluto what you call it. It is an interesting object, goes around the sun, and shows geology and an atmosphere.
There’s a tendency to define objects based on what they are now, but nothing is constant in the universe. There are some issues with the nomenclature, and a definition today may not apply to the same object tomorrow.
