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netineya [11]
3 years ago
8

Why are atoms electrically neutral?

Physics
1 answer:
Ghella [55]3 years ago
8 0
Because the electricity in the atoms are surrounded by the nucleus which is surrounded by the electron cloud
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A vibrating pendulum makes 20 complete vibrations in 12.2 seconds. Determine the vibrational period of the pendulum, along with
Hitman42 [59]

Answer:

And the same can be said of a pendulum vibrating about a fixed position or of a guitar ... along the time axis, it is possible to determine the time for one complete cycle. ... It takes 2.3 seconds to complete the first full cycle of vibration. ... Since the standard metric unit of time is the second, frequency has units of cycles/second.

Explanation:

Plz mark brainliest im only 12. and i have never been marked brainliest.

4 0
3 years ago
For your senior project, you are designing a Geiger tube for detecting radiation in the nuclear physics laboratory. This instrum
Paha777 [63]

Answer:

Maximum linear charge density = 84.14 nC/m

Explanation:

Looking at this question, The electric field of a line charge of infinite length is given by : Er = (1/(2πεo)) x (λ/r)

r = the distance from the center of the line of charge

λ = the linear charge density of the wire.

Now looking at the equatiom, due to the fact that Er varies inveresely with r, its maximum value will occur at the surface of the wire where r = R, the radius of the wire:

And so, Emax = (1/(2πεo)) x (λ/R)

Let's make λ the subject of the equation and we get;

λ = 2πεo(REmax)

From the question, R = 0.55/2 = 0.275cm

Also, Emax = 5.50 × 10^(6) N/C

Let's take the value of the electric constant to be εo = 8.854 x 10^(-9) C^(2) / Nm^2

R = 0.275mm = 0.000275m

Plugging these values into the equation, we get;

λ = 2π x 8.854 x 10^(-12) x 0.000275 x 5.50 × 10^(6) = 84.14 nC/m

4 0
3 years ago
Seasons are more noticeable in places that are____from to the equator
Liula [17]
1. farther
2. true

the equator is a weirdo
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3 years ago
What does the model below represent?
Viefleur [7K]

Answer:

D

Explanation:

Because we have two reactants and product

5 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
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