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sukhopar [10]
3 years ago
9

As our Sun exhausts its supply of hydrogen, it will cool down by 1000 to 2000 Kelvin. But it will also become about 100 times br

ighter than it is now. Describe the path our Sun will take on the HR diagram, and what type of star the Sun will become.​

Physics
1 answer:
sveta [45]3 years ago
5 0

Answer:

It will move upward and to the right, from the main sequence to the giants.

Explanation:

Plato answer so change it up a bit xx

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In general, if the temperature of a chemical reaction is increased, the reaction rateA. IncreasesB. decreasesC. remains the same
In-s [12.5K]
A. Increases

I would assume this to be the answer because heat is another form of energy. If there is more energy the molecules will become more active. This makes A the most logical answer.
6 0
3 years ago
Describe how the design of a vacuum flask keeps the liquid inside hot ?
JulijaS [17]
The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection 
3 0
3 years ago
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We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op
Kryger [21]

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒  v_{rolling}=v_{translational}+v_{rotational}

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒  \omega=\frac{v_{translational}}{r}

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒  v_{rotational}=\omega \times r

On substituting the estimated values, we get

⇒                   =\frac{v_{translational}}{r} \times r

⇒                   =v_{translational}

So that the velocity will be:

⇒  v_{rolling}=v+v

⇒              =2v

4 0
3 years ago
The potential energy of a 35 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
Anettt [7]

From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.

We need to find its height.

We will use the formula P.E = mgh

Therefore h = P.E / mg

where P.E is the potential energy,

m is mass in kg,  

g is acceleration due to gravity (9.8 m/s²)

h is the height of the object's displacement in meters.

h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters  high.

6 0
3 years ago
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