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snow_lady [41]
3 years ago
6

Your answer is incorrect. an empty swimming pool needs to be filled to the top. the pool is shaped like a cylinder with a diamet

er of 8 m and a depth of 1.6 m . suppose water is pumped into the pool at a rate of 11 m3 per hour. how many hours will it take to fill the empty pool? use the value 3.14 for π , and round your answer to the nearest hour. do not round any intermediate computations.
Physics
1 answer:
evablogger [386]3 years ago
4 0

How can my answer be incorrect if I haven't even figured it out yet ? ?

The volume of a cylinder is (pi) (radius²) (height, length, or depth).

The pool's diameter is 8m, so its radius is 4m .

The volume of the pool is (pi) (4m)² (1.6 m) = 25.6 pi m³ .

Filling it at the rate of 11 m³/hr, it will take

(25.6 pi / 11) = 7.308... hours to fill it to the brim.

Rounded to the nearest hour, that's 7 hours.

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Answer:here you goo

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What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

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