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Anna35 [415]
3 years ago
13

According to the big-bang theory how the universe might end?

Physics
1 answer:
Troyanec [42]3 years ago
3 0

Answer:

<em>If the Universe holds enough matter, including dark matter, the combined gravitational attraction of everything will gradually halt this expansion and precipitate the ultimate collapse. Over time, galaxies, then individual stars, will smash into each other more frequently, killing off any life on nearby planets.</em>

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An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

8 0
3 years ago
What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each
Arturiano [62]

\huge\mathfrak\red{✔Answer:-}

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

3 0
3 years ago
If the spring constant is doubled , what value does the period have for a mass on a spring?
zhuklara [117]

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}

So the correct answer is

D. The period would decrease by sqrt (2)

8 0
3 years ago
Read 2 more answers
what is the acceleration of each body of mass 5kg rests on a frictionless table and is connected to a cable that passes over a p
Oksanka [162]

Answer:

6.53 m/s²

Explanation:

Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.

Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:

T = m₁a               (1)

m₂g - T = m₂a    (2)

substituting T = m₁a in equation 2:

m₂g - m₁a = m₂a

m₂a + m₁a = m₂g

a(m₁ + m₂) = m₂g

a = m₂g / (m₁ + m₂)

a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²

Both objects have an acceleration of 6.53 m/s²

8 0
3 years ago
What type of force keeps the Earth in orbit around the Sun?
faust18 [17]
The answer is the suns gravity
5 0
3 years ago
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