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2 years ago

Spurs"" are probably the result of ____.

Physics

1 answer:

egoroff_w [7]2 years ago

5 0

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.

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What is the quantity of work done when a crane lifts a 100-n block from 2 m above the ground to 6 m above the ground?.

mylen [45]

Answer: by definition work = Force * distanceForce = 100Ndistance = 6 - 2 = 4work = 100 * 4 = 400 J or CStill stuck? Get 1-on-1 help from an expert tutor now.

Explanation:

7 0

2 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

3 0

3 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

3 years ago

Which will remain the same for two identical books, one lying flat and the other standing on an end?

Fiesta28 [93]

Weight will remain the same for two identical books, one lying flat and the other standing on an end.

The strain at a factor internal a liquid is at once proportional to the intensity of the factor. When an item is submerged in a liquid, the intensity of its backside from the floor of the liquid is extra than that of some other a part of the item.

Archimedes' precept is the declaration that the buoyant pressure on an item is identical to the load of the fluid displaced with the aid of using the item.

Ensure your scale is on a flat, strong and stage floor. Do now no longer use your scale on carpet. When taking measurements, stand nevertheless withinside the middle of the platform till all measurements are displayed, and if feasible do now no longer circulate your scale in-among measurements.

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8 0

1 year ago

Please help meeeee!!!!!!

Vinil7 [7]

Answer:

Net force

Explanation:

Bruh, easy question

3 0

2 years ago