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3 years ago

Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?

Physics

1 answer:

Yakvenalex [24]3 years ago

8 0

None because Neon is already stable

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

Water has a mass per mole of 18.0 g/mol, and each water molecule (H20) has 10 electrons. (a) How many electrons are there in one

Allushta [10]

Answer:

total number of electron in 1 litter is 3.34 × $10^{26}$ electron

Explanation:

given data

mass per mole = 18 g/mol

no of electron = 10

to find out

how many electron in 1 liter of water

solution

we know molecules per gram mole is 6.02 × $10^{23}$ molecules

no of moles is 1

total number of electron in water is = no of electron ×molecules per gram mole × no of moles

total number of electron in water is = 10 × 6.02 × $10^{23}$ × 1

total number of electron in water is = 6.02× $10^{24}$ electron

and

we know

mass = density × volume ..........1

here we know density of water is 1000 kg/m

and volume = 1 litter = 1 × $10^{-3}$ m³

mass of 1 litter = 1000 × 1 × $10^{-3}$

mass = 1000 g

total number of electron in 1 litter = mass of 1 litter × $\frac{molecules per gram mole}{mass per mole}$

total number of electron in 1 litter = 1000 × $\frac{6.02*10{24}}{18}$

total number of electron in 1 litter is 3.34 × $10^{26}$ electron

8 0

3 years ago

Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude = 2.20 mm

b) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{743}{2\pi}$

f = 118.25 Hz

c) wavelength

$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

$\lambda= 0.895\ m$

d) speed

$v = \dfrac{\omega}{k}$

$v = \dfrac{743}{7.02}$

v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

$T = \dfrac{v^2\ m}{L}$

$T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}$

T = 27.87 N

g) Power transmitted by the wave

$P = \dfrac{1}{2}m\ v \omega^2\ A^2$

$P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2$

P = 0.438 W

5 0

3 years ago

A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p

MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) $f=1617.9\ Hz$

(b) $T=0.618\ ms$

(d) $\varphi=60^o$

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

$V(t)&=A\cdot \sin(2\pi ft+\varphi )$

Where

$A=Amplitude$

$f=frequency$

$\varphi=Phase\ angle$

The voltage of the question is the sinusoid expression

$V(t)=20cos(5\pi\times 103t+60^o)$

(a) By comparing with the general formula we have

$f=5\pi\times 103=1617.9\ Hz$

$\boxed{f=1617.9\ Hz}$

(b) The period is the reciprocal of the frequency:

$\displaystyle T=\frac{1}{f}$

$\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec$

Converting to milliseconds

$\boxed{T=0.618\ ms}$

$\boxed{A=20 \ Volts}$

(d) Phase angle:

$\boxed{\varphi=60^o}$

4 0

2 years ago

Please list the following for the Element, HYDROGEN

bagirrra123 [75]

Answer:

1 Proton, 1 Electron, No Neutrons

Group 1, Period 1

Gases

3 0

2 years ago