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GaryK [48]
3 years ago
14

The addition of electron shells results in shielding of electrons from the nucleus.

Physics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

Addition of shells increases the distance of outer electrons from the nucleus.

Explanation:

Shielding effect is known as the attraction between the nucleus and an electron of any atom. In other words, it is the reduction in effective nuclear charge on an electron cloud.

Addition of electron shells results in the shielding of electron from nucleus.  As the number of electron shells increases then farther will be the electrons placed from the nucleus and hence it will become easier to expel the electrons from outer shells with only little amount of ionization energy.

So, the amount of ionization energy require will be indirectly proportional to the shielding effect because more the shielding of electrons from the nucleus less will be the ionization energy require to expel the electrons.

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On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced
Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

v =f\lambda

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

\mu = \frac{m}{l}

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m

Now, for the tension we use the formula:

v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0
3 years ago
A motorbike reaches a speed of 20 m/s over 60m, whilst
Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation  </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration  

An equation of uniformly accelerated motion  

V = vo + at  

Vt² = vo² + 2a (x-xo)  

x = distance on t  

vo / vi = initial speed  

vt / vf = speed on t / final speed  

a = acceleration  

vf=20 m/s

d = 60 m

a = 3 m/s²

\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s

7 0
3 years ago
Which statement BEST describes Miguel Hidalgo?
tatiyna
Miguel Hidalgo was a Priest so I think it would be 1. 
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6 0
3 years ago
Read 2 more answers
An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH
Doss [256]

Answer:

Xc= 17.267 Ω,   Z= 415.5 Ω,   I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH,  C = 23.1 μF

a)

Capacitive reactance = Xc=?

Xc= \frac{1}{2\pi fC}

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z=\sqrt{ R^2 + (Xl - Xc)^2}

Xl= 2π * f * L  

Xl= 2π * 399 * 147 * 10^{-3}

Xl= 368.5 Ω

Z=\sqrt{ R^2 + (Xl - Xc)^2} = \sqrt{222^2 + (368.5 - 17.267)^2}

Z= 415.5 Ω  

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0
3 years ago
A light wave has a wavelength of 450 nanometers. What is the frequency of this light?
Zolol [24]

Answer:

Frequency, f=6.67\times 10^{14}\ Hz

Explanation:

Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,

c=f\lambda

f is the frequency of this light

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{450\times 10^{-9}}\\\\f=6.67\times 10^{14}\ Hz

So, the frequency of this light is 6.67\times 10^{14}\ Hz.

3 0
3 years ago
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