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4 years ago

10

Will an atom emit light if all the atoms electrons are in the ground state

1 answer:

Oliga [24]4 years ago

6 0

No,

To emit light an electron has to jump down to a lower energy level but in an electron is unable to go any lower in ground state.

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Why do Kepler's laws describe?

leva [86]

Answer:

The shape of an orbit

Explanation:

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3 years ago

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A rectangular pane of glass is 91.1 cm wide and 155.9 cm long, and its area is equal to the length multiplied by the width. Usin

Kamila [148]

911.1×155.9 = 14,202.49
14,202.49 in 3 significant figures would be either 14,200 or 1.42×10^4

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3 years ago

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Why does the heat of the heating element stop when the desired temperature in an electric iron is reached?

Natalija [7]

The heat of the heating element stops when the desired temperature in an electric iron is reached because A. the thermostat blade opens electrical contacts.
This is why heating ceases to rise or fall (it stops completely) once the thermostat's blade opens.

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3 years ago

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Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.

Vsevolod [243]

Answer:

The angular speed of the neutron star is 3130.5 rad/s.

Explanation:

Given that,

Initial radius $r_{1}=7\times10^{5}\ km$

Final radius $r_{2}=18 km$

Density of a neutron $\rho= 10^{14}$

Equal masses of two stars $m_{1}=m_{2}$

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T = 35 days = 3024000 s

We need to calculate the initial angular speed of original star

Using formula of angular star

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

Put the value into the formula

$\omega_{1}=\dfrac{2\pi}{3024000}$

$\omega_{1}=0.00207\times10^{-3}\ rad/s$

Let the initial moment of inertia of the star is

$I_{1}=m_{1}r_{1}^2$

Final moment of inertia of the star is

$I_{2}=m_{2}r_{2}^2$

From the conservation of angular momentum

$I_{1}\omega_{1}=I_{2}\omega_{2}$

$\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}$

$\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}$

Put the value into the formula

$\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}$

$\omega_{2}=3130.5\ rad/s$

Hence, The angular speed of the neutron star is 3130.5 rad/s.

8 0

3 years ago

A very long insulating cylinder of charge of radius 2.60 cm carries a uniform linear density of 15.0 nC/m . Part A If you put on

fenix001 [56]

Given Information:

Radius = ra = 2.60 cm = 0.026 m

Density = J = 15.0 nC/m

change in potential difference = ΔV = 200 V

Required Information:

Distance = d = ?

Answer:

distance = 0.088 m

Explanation:

As we know

ΔV = Vb - Va = J/4πε₀*ln(rb/ra)

Where ra and rb is the point where potential difference is Va and Vb respectively

1/4πε₀ = 9x10⁹ N.m²/C²

We want to find the distance d = rb - ra

ΔV = J/4πε₀*ln(rb/ra)

200 = 9x10⁹*15x10⁻⁹*ln(rb/ra)

200/135 = ln(rb/ra)

1.48 = ln(rb/ra)

taking e on both sides yields

e^(1.48) = rb/ra

4.39 = rb/ra

rb = 4.39*0.026

rb = 0.114 m

Therefore, the required distance is

d = rb - ra

d = 0.114 - 0.026

d = 0.088 m

Therefore, the other probe must be placed 0.088 m from the surface so that the voltmeter reads 200 V

6 0

3 years ago