Spurs are probably the result of <u>self-sustaining</u> <u>star formation.</u>
<h3>What is the formation of gaseous spurs in spiral galaxies?</h3>
The gigantic form of the magnificent doppelganger spiral patterns that spiral outward from the galactic cores gave spiral galaxies their name. These light arms of spiral galaxies are frequently seen in optical pictures to be speckled with bright star-forming areas at regular intervals.
Smaller structures spread forth and rearward into the interarm area from each major spiral arm. Spiral-arm also known as spurs are the name given to these substructures. Sometimes the spurs are also filled with star-forming clusters. As a consequence, we may draw the conclusion that spurs most likely emerge from self-sustaining star formation.
Learn more about the spiral galaxies here:
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Here's a useful factoid that you don't hear about very often:
1 volt means 1 Joule per Coulomb.
When 1 coulomb of charge falls or gets lifted through 1 volt potential difference, it gains or loses 1 Joule of energy.
If you want to lift 5 coulombs to a height of 1 volt, you have to give it 5 joules.
If you actually give those 5 coulombs 7.5 joules instead, they'll rise up to 1.5 volts above the potential where they started. The flowed through a potential DIFFERENCE of 1.5 volts.
(If they started at a point that's connected to the Earth, like a water pipe or a metal flagpole, then their new potential is 1.5 volts, because we define zero as the potential of the ground.)
X =(3.00x4.00 x3-1.00t x 2.00) x m
x= (12.00x3- 1.00 x2.00) x m
x= 36.00 -1.00 x 2.00) x m
x = (36.00 -2.00) x m
x =( 34.00) x m
x =34.00 times m
Answer:
The work done is 205 kJ.
Explanation:
Hi there!
Work can be calculated using the following equation:
W = F · Δx
Where:
W = work
F = applied force
Δx = displacement
In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:
W = ∫ F · dx
F = 3.6 N/m³ · x³ - 76 N
W = ∫ (3.6 x³ - 76)dx
W = 0.9 x⁴ - 76x
Evaluating from xi to xf:
W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m
W = 205 kJ
