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3 years ago

How are stars distributed throughout an elliptical galaxy?

1 answer:

6 0

An elliptical galaxy is a type of galaxy having an approximately ellipsoidal shape and a smooth, nearly featureless brightness profile. Unlike flat spiral galaxies with organization and structure, they are more three-dimensional, without much structure, and their stars are in somewhat random orbits around the center.

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Atoms that are bonded together to form a new material with new properties and characteristics. .

mihalych1998 [28]

Answer:

Explanation:

Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

7 0

2 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -

Liula [17]

Answer:

Electric field, $E=6.86\times 10^6\ N/C$

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, $q=-3\ nC=-3\times 10^{-9}\ C$

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

$ma=qE$

a = g

$E=\dfrac{mg}{q}$

$E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}$

$E=6860000\ N/C$

$E=6.86\times 10^6\ N/C$

Hence, the magnitude of electric field that balances its weight is $6.86\times 10^6\ N/C$ . Hence, this is the required solution.

4 0

3 years ago

La capital de Puerto Rico es _____.

velikii [3]

La capital de Puerto Rico es la ciudad "San Juan".

7 0

3 years ago

Read 2 more answers

A pendulum is timed, first for 20 swings and then for

wlad13 [49]

$w = \frac{t(50) - t(20)}{50 - 20} = \frac{43.2 - 17.4}{50 - 20}$

$w = \frac{25.8}{30} = 0.86 \: sec \: per \: swing$

The results may differ due to resistive forces that may be affecting the system by decelerating it or any other external forces that might accelerate it a bit.Or the timing could be a little inaccurate.

3 0

1 year ago