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Cerrena [4.2K]
2 years ago
15

What is the order of magnitude of the distance of Sun to nearest star in meters?

Physics
2 answers:
Brilliant_brown [7]2 years ago
6 0

Answer:

Mercury, 46,001,272 km from the sun at the nearest point.

Explanation:

neonofarm [45]2 years ago
4 0

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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7nadin3 [17]

Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

F = 351×10³lb.

4 0
3 years ago
Rosa pours a cup of boiling water into a pot of room-temperature water. According to the second law of thermodynamics, what will
11111nata11111 [884]

<u>Thermal energy</u><u> from the room-temperature water will continuously flow to the boiling water.</u>

  • The second law states, in a straightforward manner, that heat cannot naturally go "uphill."
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8 0
1 year ago
a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all dire
My name is Ann [436]

Answer:

I=2.67\times 10^{-6}\ W/m

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m

So, the wave intensity 25 km from the antenna is 2.67\times 10^{-6}\ W/m.

5 0
3 years ago
Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a
Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

P = \frac{F}{A}

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

P = \frac{40\ N}{0.03\ m^2}\\\\

<u>P = 1333.33 N</u>

4 0
3 years ago
A. In your own words, define what a transverse wave is.
olasank [31]

Answer:

a) From definition a transverse wave is which one where the elements moves perpendicular to the direction of the wave. For example is a wave is moving from the left to the right the elements would be wibrating or moving upward or downward.

We have a lot examples for a transverse wave. For example water waves, strings on the musical instruments , light and radio waves.

b) We can identify a transverse wave if the particles are displaced perpendicular to the direction of the wave. Usually these types of wave occur in elastic solids. And we can identify it when we see a pattern perpendicular between the wave direction and the particles motion. In simple words we need to see that the wave is moving down and up.

Explanation:

Part a

From definition a transverse wave is which one where the elements moves perpendicular to the direction of the wave. For example is a wave is moving from the left to the right the elements would be wibrating or moving upward or downward.

We have a lot examples for a transverse wave. For example water waves, strings on the musical instruments , light and radio waves.

Part b

We can identify a transverse wave if the particles are displaced perpendicular to the direction of the wave. Usually these types of wave occur in elastic solids. And we can identify it when we see a pattern perpendicular between the wave direction and the particles motion. In simple words we need to see that the wave is moving down and up.

3 0
3 years ago
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