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Artemon [7]
3 years ago
6

Helpppppp please anybody who knows

Physics
1 answer:
levacccp [35]3 years ago
3 0
We know that the intensity of light decreases as the square of the distance
from the source.

Lamp-1 is twice (2x) as far from the detector as Lamp-2 is.
If they appear equal to the detector, then Lamp-1 must be (2)² = 4 times
as bright as Lamp-2.

4 x 15 candelas  =  60 candelas  


Question #15).
A). is true, but has nothing to do with particle behavior
B). Diffraction is a purely wave thing, so it doesn't help.
C). is true, but doesn't say anything about particle behavior
D). is true, but isn't an indication of particle behavior.

Looks to me like NONE of the choices is any help
to suggest particle behavior.
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3 years ago
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.

4 0
2 years ago
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Describe how can two or more velocities be combined
eduard
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