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Which coordinate system would be most useful for two observers (one in Michigan and one in Florida) who wants to observe the sam

Mkey [24]

Answer : Celestial or azimuth - altitude

Explanation :

Celestial : The celestial coordinates that are analogous to longitude and latitude are called RA and Dec.

RA = Right Ascension

Dec = Declination

RA is the measured in unit of time and Dec is measured in degree.

The equatorial coordinate system is the projection of the latitude and longitude coordinate system on the celestial sphere.

Azimuth - altitude : Azimuth - altitude define the location of an object in the sky.

The altitude is the distance of an object appears to be above the horizon.

The Azimuth of the object is the angular distance along the horizon to the location of the object.

3 0

3 years ago

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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.

4 0

3 years ago

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In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.

nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

$p=mv$

$p=25kg*3m/s= 75kg*m/s$

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

$75kg*m/s = 5.0kg*v$

$v=\dfrac{75}{5}$

$\boxed{v=15m/s}$

8 0

3 years ago

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to

goldenfox [79]

Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

$T$ = tension force in the tow rope = 350 N

$L$ = length of the incline surface = 120 m

$W_{t}$ = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

$W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J$

$h$ = height gained by the rider = 30 m

$m$ = total mass of rider and tube = 84 kg

Potential energy gained is given as

$U = mgh\\U = (84) (9.8) (30)\\U = 24696 J$

$Q$ = Thermal energy created

Using conservation of energy

$Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J$

7 0

3 years ago

How much tension must a rope withstand if used to accelerate a 960-kg car from rest horizontally along a frictionless surface to

Svetach [21]

Use a=(dv/dt) (change in velocity/ change in time)=acceleration
(1.2/5)=acceleration

F=ma (Newton's second law, Force= Mass x Acceleration

=960 x 0.24 F=230.4N If T<230.4N then the tow rope will hold

7 0

3 years ago