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3 years ago

6

What do halo stars do differently from disk stars?

1 answer:

VLD [36.1K]3 years ago

6 0

Answer: They orbit the galactic center with many different inclinations, while disk stars all orbit in nearly the same plane. ... They have vertical motions out of the plane, making them appear to bob up and down, but they never get "too far" from the disk.

Explanation:

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Describe how Ubuntu could help to lack of basic services challenges<br>

Butoxors [25]

Explanation:

Ubuntu is somewhat a South African concept that involves charity, sympathy, and mainly underlines the concept of universal brotherhood. Hence this concept can help fight social challenges such as racism, crime, violence and many more. It can contribute to maintaining peace and harmony in the country at large

4 0

3 years ago

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Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh

Nutka1998 [239]

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

$a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s$

Let d is the distance cover during that time. So,

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m$

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

4 0

3 years ago

Anyone know the answer ?

nata0808 [166]

Answer:

d

Explanation:

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3 years ago

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The rate (in cubic feet per hour) that a spherical snowball melts is proportional to the snowball's volume raised to the 2/3 pow

Darina [25.2K]

Answer:

A 3 feet radius snowball will melt in 54 hours.

Explanation:

As we can assume that the rate of snowball takes to melt is proportional to the surface area, then the rate for a 3 feet radius will be:

T= A(3 ft)/A(1 ft) * 6 hr

A is the area of the snowballs. For a spherical geometry is computing as:

A=4.pi.R^2

Then dividing the areas:

A(3 feet)/A(1 foot) = (4 pi (3 ft)^2)/(4 pi (1 ft)^2) = (36pi ft^2)/(4pi ft^2)= 9

Finally, the rate for the 3 feet radius snowball is:

T= 9 * 6 hr = 54 hr

6 0

3 years ago

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time

Makovka662 [10]

Explanation:

We have,

Surface area, $A=2\ cm^2=0.0002\ m^2$

The current varies wrt time t as :

$q(t) = 4t^3 + 5t + 6$

(a) At t = 2 seconds, electrical charge is given by :

$q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C$

(b) Current is given by :

$I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5$

Instantaneous current at t = 1 s is,

$I=12(1)^2+5=17\ A$

(c) Current is, $I=12t^2+5$

Current density is given by electric current per unit area.

$J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2$

Therefore, it is the required explanation.

7 0

4 years ago