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ipn [44]
3 years ago
11

Which of the following wire gage sizes is the thickest? A. 8 B. 10 C. 14 D. 0

Physics
1 answer:
maks197457 [2]3 years ago
5 0
A because it's the smaller the thicker you just can't have 0 gage
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What pushes against gravity in: a main sequence star, a white dwarf, a neutron star, and a black hole? electron degeneracy, neut
Inga [223]

Answer:

heat pressure, electron degeneracy, neutron degeneracy, and nothing

Explanation:

Main Sequence Star: It is a star in which nuclear fusion is happening in the core of the star. Hydrogen molecules fuse together to generate Helium. This nuclear fusion generates outward gas pressure and radiation pressure which balances the inward gravity thus creating an equilibrium which keeps the stars in shape.

White dwarf: It is the end stage of a medium sized star like the Sun. Outer layers of the star are thrown in the form a shell/bubble leaving a small and dense core in the center called as white dwarf. This core consists of carbon and oxygen. Nuclear fusion doesn't occur in the core of white dwarfs. The inward gravity is balanced by the electron degeneracy pressure. Thus these stars will keep on radiating the remaining heat and will turn in to a black dwarf at the end.

Neutron Star: This is the end stage of a supermassive star (1-3 times the mass of the Sun). At the last stage of the life the core collapses. In these stars the inward gravity is so huge that the pressure overcomes the electron degeneracy pressure and crushes together the electron and proton to form neutron. The neutron then stops the collapse and balances the inward gravity.

Black Hole: This is the end stage of a hyper massive stars weighing more than 3 times the mass of the Sun. The inward gravitational force is so huge that even the neutrons are not able to stop the collapse the core. thus the mass of the star collapses into a very small area of immense gravity. There is nothing that can balance this inward gravity.

3 0
2 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

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The answer is: a. Skinner’s idea of language

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3 years ago
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