The star with apparent magnitude 2 is more brighter than 7.
To find the answer, we have to know about apparent magnitude.
<h3>What is apparent magnitude?</h3>
- 100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
- The apparent magnitude of bigger stars is always smaller.
- The brightest star in the night sky is Sirius.
- The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
- The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.
Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.
Learn more about the apparent magnitude here:
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Force , F = ma
F = m(v - u)/t
Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.
m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s
F = 1.2*10³(20 - 10)/5
F = 2.4*10³ N = 2400 N
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.
Ok, Rameses:
Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton
Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg
a). The force between two charges is
F = (9 x 10⁹) Q₁ Q₂ / R²
= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²
= ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²
= 8.05 x 10⁻⁸ Newton .
b). Centripetal acceleration =
v² / r .
A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)
= 7.7 x 10²² m/s² .
That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.
It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.
I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .
I'm going to leave that step for you. Good luck !
Answer:
600m
Explanation:
30×20 at a constant speed is 600m.
