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lara [203]
4 years ago
10

I WILL MARK YOU AS BRAINLIEST IF RIGHT

Physics
1 answer:
butalik [34]4 years ago
6 0

Answer:

25.4 miles per hour

Explanation:

speed=distance/time

speen=524/20.6 hours

=25.4 miles per hour

You might be interested in
Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.
nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

  • The kinetic energy ratio is   \frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2
<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

          K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

  • Translational kinetic energy. Due to the linear movement of the body

            K_{tras} =\frac{1}{2} m v^2

  • Rotational kinetic energy. Due to the rotational movement of the body.

            K_{rot} = \frac{1}{2} I w^2

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

           K_{earth} = K_{tras} + K_{rot}

Linear and rotational speed are related.

           v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

           I = \frac{2}{5 }  m r^2  

Let's  subatitute.

         

          K_{earth} = \frac{1}{2} \  m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}  

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

       w = \frac{2 \pi}{T}  

Let's substitute.

        K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2  \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2  

        K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)  

Data for Earth are tabulated:

  • Mass m = 5.98 1024 kg
  • Radius r = 6.37 10⁶ m
  • Radius orbits tras = 1.496 10¹¹ m
  • Rotation period T_{rot} = 24 h (\frac{3600s}{1h}) = 8.64 10⁴s
  • Translation period  T_{tras} = 365 d (\frac{24h}{1 d}) (\frac{3600s}{1h}) = 3.15 10⁷ s

Let's calculate.

        K_{earth} = 4 \pi^2 5.98 \ 10^{24}  ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2  \ +  \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )

        K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)

        K_{earth}= 2.66 \ 10^{33} J

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of T_{Sum} = 27 days.

The kinetic energy of the sun is;

          K_{sum} = K_{rot} =  \frac{1}{2} I w^2  

          K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2  

          K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2  

The tabulated data for the sun are:

  • Mass m = 1,991 1030 kg.
  • Radius R = 6.96 10⁸ m
  • Period T = 27 d (\frac{24h}{1 d} ) (\frac{3600s}{1h}) = 2.33 10⁶ s

         

Let's calculate.

           

          K_{sum} = 1.40 \ 10^{36} J

The relationship of the kinetic energy of the sun and the Earth is:

        \frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}  

       \frac{K_{sum}}{K_{earth}} =  5.3 \ 10^2  

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

  • The kinetic energy ratio is:  \frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2

Learn more about kinetic energy here: brainly.com/question/25959744

5 0
3 years ago
Why are solar winds potentially fatal for organisms living on earth?
Amiraneli [1.4K]
Solar winds contain high levels of radiation which could lead to various health related complications in living organisms on Earth. These complications include and are not limited to radiation sickness, chromosomal damage, and cancer. However, solar winds are blocked by the Earth's magnetic field, only posing problems to our telecommunications.  
7 0
4 years ago
Erickson’s psychosocial stages emphasize
Musya8 [376]
This question is true because it just is
4 0
3 years ago
Write down the addres of earth in as much detail as possible.
solong [7]
3rd nearest planet.
Neighborhood of Sun.
Orion arm.
Milky Way galaxy.
Local cluster.
13.8 billion years.
Universe.
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8 0
3 years ago
Calculate the hydrostatic difference in blood pressure between the brain and the foot in a person of height 1.93 m. The density
Slav-nsk [51]

Answer:

Explanation:

Given: Density of blood = 1.03 × 10³ Kg/m³, Height =  1.93 m g = 9.8 m/s²

pressure at the brain is equal to atmospheric pressure. = Hydro-static

pressure(ρ₀)

∴ pressure of the foot = pressure of the brain(ρ₀) + ( density of blood × acceleration due to gravity × height)(ρgh)

Hydro-static pressure = pressure at the feet- pressure at the brain(ρ₀)

Hydro-static pressure (Δp) = (ρgh + ρ₀) - ρ₀ = ρgh

Hydro-static pressure = 1.03 × 10³ × 9.8 × 1.93 = 1.948 × 10⁴ Pa

∴  Hydro-static pressure ≈ 1.95 × 10⁴ Pa

3 0
3 years ago
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