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3 years ago

What happens when an electron moves from an excited state to the ground state?

Physics

1 answer:

Darina [25.2K]3 years ago

6 0

<span>When an electron moves from an excited state to the ground state, "Energy releases"

Hope this helps!</span>

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Three equal point charges, each with charge 2.00 μC , are placed at the vertices of an equilateral triangle whose sides are of l

oksano4ka [1.4K]

Answer:

Explanation:

Energy of system of charges

= k q₁q₂ / r₁₂ + k q₁q₃ / r₁₃ + k q₃q₂ / r₃₂

q₁ , q₂ and q₃ are charges and r₁₂ , r₁₃ , r₃₂ are densities between them

9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )

= 9 x 10⁹ x 3 x 16 x 10⁻¹²

= 432 x 10⁻³

= .432 J .

4 0

3 years ago

3. The electric field of a sinusoidal electromagnetic wave has an amplitude of 5.0 V/m. How much radiation energy passes through

True [87]

Answer:

e) $179$ J

Explanation:

$E_{o}$ = Magnitude of electric field = 5 V/m

$A$ = Area of window = 1.5 m²

$c$ = speed of electromagnetic wave = 3 x 10⁸ m/s

$\Delta t$ = time interval = 1 h = 3600 sec

radiation energy is given as

$U = (0.5)\epsilon _{o}E_{o}^{2}cA\Delta t$

$U = (0.5)(8.85\times 10^{-12})(5)^{2}(3\times 10^{8})(1.5)(3600)$

$U = 179$ J

6 0

3 years ago

A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball

RideAnS [48]

Answer:

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, $F_r=Cv^2$

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

$Cv^2=mg$

$C=\dfrac{mg}{v^2}$

$C=\dfrac{0.046\times 9.8}{(44)^2}$

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Hence, this is the required solution.

4 0

3 years ago

I’m not sure how to solve this

spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0

3 years ago

A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag

ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

$B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}$

$B=10^{-7}\times \frac{2\times140}{5}$

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0

3 years ago