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anygoal [31]
3 years ago
9

What is the frequency of radiation emitted by a photon of light if the energy released during its transition to ground state is

3.611 × 10-15 joules? (Planck's constant is 6.626 × 10-34 joule seconds) 5.50 × 10-11 Hz 2.39 × 10-50 Hz 5.45 × 1018 Hz 1.633 × 1027 Hz
Physics
1 answer:
miskamm [114]3 years ago
3 0
If i was feeling harsh today, I'd say the answer to your question is impossible to obtain due to the fact that photons do not emit radiation, photons ARE the radiation emitted. Though for the sake of it, here is the method...

<u>The simple method:
</u>

E=hf

therefore f=e/h

f=(3.611x10^-15) / 6.63x10^-34)

Answer: 5.45x10^18
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REY [17]

The magnitude of the charge on the balloon is 1.6 x 10⁻¹² C.

<h3>What is the magnitude of the charge on the ball?</h3>

The magnitude of the charge on the ball is calculated by determining the total charge equivalent to the given number of electrons.

The charge of one electron = 1.6 x 10⁻¹⁹ Coulombs

Now, we are going to estimated the total charge of 1 x 10⁷ electrons.

1 electron =  1.6 x 10⁻¹⁹ C

1 x 10⁷ electrons = ?

= (1 x 10⁷ electrons x 1.6 x 10⁻¹⁹ C) / (1 electron)

= 1.6 x 10⁻¹² C

Thus, the total charge of 1 x 10⁷ electrons is obtained by multiplying the magnitude of charge of one electron to the entire given electrons.

Learn more about charge of electron here: brainly.com/question/9317875

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8 0
1 year ago
How much heat in joules is required to heat a 50 g sample of aluminum from 71 ∘f to 142 ∘f?
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3 years ago
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Answer:

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3 years ago
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular
Evgesh-ka [11]

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

2\alpha \theta=\omega_f^2-\omega_i^2

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}

<u>α = - 1930.2 rad/s²</u>

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5 0
3 years ago
A 1.53-kg bucket hangs on a rope wrapped around a pulley of mass 7.07 kg and radius 66 cm. This pulley is frictionless in its ax
levacccp [35]

Answer:

\alpha = 6.431\,\frac{rad}{s^{2}}

Explanation:

The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

\Sigma M = T \cdot R = \frac{1}{2}\cdot M \cdot R^{2}\cdot \alpha

Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

T = \frac{1}{2}\cdot M \cdot R \cdot \alpha

m_{b}\cdot g = \frac{1}{2}\cdot M \cdot R \cdot \alpha

\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}

\alpha = \frac{2\cdot (1.53\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(7.07\,kg)\cdot (0.66\,m)}

\alpha = 6.431\,\frac{rad}{s^{2}}

7 0
3 years ago
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