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SpyIntel [72]
3 years ago
6

What is a limitation of the Big Bang model?

Physics
1 answer:
bogdanovich [222]3 years ago
5 0
The size and the explosion
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If sulfur 34 undergoes alpha decay, what will I become?
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It will decay into Silicon-30. Because alpha particles are 2 protons and 2 neutrons with an atomic mass of 4, you minus sulfur's atomic number by 2 and get silicon. And the atomic mass is 34 - 4 which equals 30.
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What is the advantage of having different shaped beaks? Question 4 options: To protect itself against prey. To be able to eat th
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To be able to eat the food readily available in the environment

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Kate gathered 3 of the same size made of different materials: glas, clear plastic, and aluminum painted black. She placed on the
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<u>Answer:</u>

<em>To relate the type of box material to the warmth inside the box </em>

<u>Explanation:</u>

3 boxes are made with three different materials glass, plastic and Aluminium. Thermal conductivity is different for different materials. thermal conductors allow easy flow of heat through them and insulators allow minimal or no flow of heat through them.

Thus the amount by which an object gets heated up depends on the value of its conductivity. In this experiment, glass and plastic are insulators and aluminium is a conductor. Among glass and plastic, plastic is a better insulator.

<em>Thus the heat contained in the boxes will be of the order </em>Aluminium>glass>plastic.

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3 years ago
Which division of time on the geologic time scale is the shortest?
Scrat [10]

Explanation:

In geology, there are 5 divisions of time on the time scale. They are:

  1. eons
  2. eras
  3. periods
  4. epochs
  5. ages

Eons are the largest time period, while ages are the shortest time period. The rest of the above listed are in between the two.

8 0
3 years ago
A conducting spherical shell with inner radius a and outer radius b has a positive point charge +Q at the center in the empty re
Andreas93 [3]

Answer:

a) q_inner = -Q

, q_outer = -2Q

b)    E₁ = k Q / r²        r<a

       E₂ = 0               a<r<b

       E₃ = - k 2Q/r²     r>b

d)   the charge continues inside the spherical shell, the results do not change

Explanation:

a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign  

q_inner = -Q  

the outer shell of the shell the load is  

q_outer = -3 Q + Q  

q_outer = -2Q

b) To find the electric field again, use Gauss's law,  

We define as a Gaussian surface a sphere  

Ф = E. dA = q_{int}/ε₀

in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product  

E A = q_{int}/ε₀

the area of ​​a sphere is  

A = 4 π r²  

E = 1 / 4πε₀  Q/ r²  

k = 1 / 4πε₀  

let's apply this expression to the different radii

i) r <a  

in this case the load inside is the point load  

q_{int}= + Q  

E₁ = k Q / r²

ii) the field inside the shell  

a <r <b  

As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.  

E₂ = 0

iii)  r>b

   q_{int}  = Q- 3Q = -2Q

    E₃ = k (-2Q/r²)

     E₃ = - k 2Q/r²

c) see  attached

d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface

6 0
3 years ago
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