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3 years ago

How do you determine the number of electrons in the shells

Physics

1 answer:

snow_tiger [21]3 years ago

4 0

Periodic table

Find your element

Locate the element atomic number

Determine the number of electrons

Look for the atomic mass of the element

subtract the atomic number from the atomic number

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PLEASE HELP I NEED THIS ANSWER BADLY!!!!!!!! Explain how might an animal respond to stimuli as compared to a plant?

Misha Larkins [42]

Explanation:

Animas and plants react to stimuli which may come from other living things or from the environment. A stimulus causes the organism which receivs it to respond to it. Therefore the stimulus changes the behavior in the organism in some way.

5 0

3 years ago

The law of conservation of energy states that

IgorLugansk [536]

Energy is neither created nor destroyed

7 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

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

The proton mass is equal to the antiproton mass, so:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$

$K_{ap} = 80.75 MeV$

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0

3 years ago

A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of

musickatia [10]

The frictional force is given by F = μmg

where μ is the coeficient of friction. 

Work done by frictional force = Fd = μmgd 

Kinetic energy "lost" = 1/2 mv² 

Fd = μmgd = 1/2 mv² 

The m's cancel μgd = v² / 2 

d = v² / 2μg 

d = 8² / 2(0.41)(9.8) 

d = 32 / (0.41)(9.8) 

d = 7.96 

Player slides 8 m . 

Note. In your other example μ = 0.46 and v = 4 m/s 

d = v² / 2μg 

= 4² / 2(0.46)(9.8) 

= 8 / (0.46)(9.8) 

= 1.77 or 1.8 m.

Hope i Helped :D

3 0

3 years ago

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