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bekas [8.4K]
3 years ago
11

Is the universe endless, how and why?

Physics
2 answers:
Grace [21]3 years ago
6 0

Answer:

Unknown

Explanation:

Because we cannot observe space beyond the edge of the observable universe, it is unknown whether the size of the Universe in its totality is finite or infinite.

Nadya [2.5K]3 years ago
4 0

The given statement is true, that is, the universe is endless due to the fact that it is not limited with a specified boundary.

<u>Explanation: </u>

Universe has unimaginable length and width due to the limit of the present day science and technological adaptation because of which we haven’t explored the full universe yet. The world in which we live is the Earth which has been cartographed and studied well along with the Milky Way galaxies.  

But beyond this, there are thousands and millions of galaxies which has not been studied. Then there is a frame of reference of study with the dimensional frame where we have been adept to 3 dimensions only and the 4^{th} dimension is yet to be studied.

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Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
Describe some ways that industry and agriculture use physical properties to separate substances.
musickatia [10]
This is more chemistry. But it is a process called fractional distillation, and it basically separates the long chained hydrocarbons from the short chained hydrocarbons through separation dependant on the boiling point of the crude oil.
3 0
3 years ago
Read 2 more answers
Does gravity increase or decrease with greater distance?
Karo-lina-s [1.5K]
Decrease. Gravity is stronger here on earth than on the moon.
3 0
3 years ago
Which data table describes the Sun? A. Composition mostly frozen nitrogen Mass 258.9 × 1018 kg B. Composition mostly hydrogen an
marshall27 [118]

The Sun is a star around which our planet Earth and other planets and satellites of our solar system revolve along specific orbits. Sun appears to be spherical in shape and consists of glowing hot gases. It contains mostly hydrogen and helium gases. The temperature of the gas mixture is extremely high.

Therefore, the Sun mostly composed of hydrogen and helium gases.

(Answer) B. Composition mostly hydrogen and helium gas, Mass 1.989 x 10^30 Kg

7 0
4 years ago
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