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Agata [3.3K]
4 years ago
8

oxygen (o) is a gas found in the 16th column of the periodic table. What statement is true about oxygen and he other elements in

that column?
Physics
1 answer:
Tpy6a [65]4 years ago
8 0
<h3>I'm pretty sure, all the elements in the column have similar chemical properties.</h3>
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A runner runs around the track consisting of two parallel lines 96 m long connected at the ends by two semi circles with a radiu
Zielflug [23.3K]
-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
7 0
3 years ago
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
g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria a
tangare [24]

Answer:

86.4 hrs

Explanation:

The amount of bacteria is initially 1

It doubles every 24 hrs.

After first 24 hrs, the amount = 2

After next 24 hrs = 4

After next 24 hrs = 8

After next 24 hrs = 16

After next 24 hrs = 32

After next 24 hrs = 64

After next 24 hrs = 128

After next 24 hrs = 256

Total time taken to reach 256 = 24 x 8 = 192 hrs

For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship

t = t'(1 - ¥^2)^0.5

Where t is the contracted time =?

t' is the time on earth

¥ = v/c

Where v is the speed of the rocket

c is the speed of light

since v = 0.893c

¥ = 0.893

Substituting, we have

t = 192 x (1 - 0.893^2)^0.5

t = 192 x 0.2025^0.5

t = 192 x 0.45 = 86.4 hrs

8 0
3 years ago
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled? *
fomenos

Answer:

B)60km[W]

Explanation:

The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].

Hope that helps!

3 0
3 years ago
Read 2 more answers
How does a laser work?
a_sh-v [17]

Answer:

A laser is created when the electrons in atoms in special glasses, crystals, or gases absorb energy from an electrical current or another laser and become “excited.” The excited electrons move from a lower-energy orbit to a higher-energy orbit around the atom's nucleus.

7 0
4 years ago
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