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WINSTONCH [101]
3 years ago
11

What is the Ultraviolet Catastrophe?​

Physics
1 answer:
Sholpan [36]3 years ago
7 0

The ultraviolet catastrophe was the prediction of late 19th century/early 20th century classical physics that an ideal black body (also blackbody) at thermal equilibrium will emit radiation in all frequency ranges, emitting more energy as the frequency increases.

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Ben Rushin is waiting at a stop light. Turns green, ben accelerated from rest at a rate of 6.00 m/s squared for a time of 4.10 s
Lera25 [3.4K]
D= vt +.5at^2
since he started at rest, v (initial velocity) is 0
so d=.5at^2
d = .5 (6m/s^2) (4.1s)^2 
then put that into a calculator.

4 0
2 years ago
An electric generator transforms mechanical energy into electrical energy. This process could be done by which of these?
Dimas [21]

Answer:

its d

Explanation:

6 0
3 years ago
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Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
Which claim would most likely be considered valid?
Mekhanik [1.2K]
Please state the options and I will answer to the best of my abilities XD
7 0
3 years ago
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An electron in a mercury atom drops
aksik [14]

Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.

<h3>How to calculate the photon energy?</h3>

In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.

Mathematically, the Planck-Einstein equation can be calculated by using this formula:

E = hf

<u>Where:</u>

  • h is Planck constant.
  • f is photon frequency.

In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.

Read more on photons here: brainly.com/question/9655595

#SPJ1

4 0
2 years ago
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