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dalvyx [7]
3 years ago
8

The "opposite" of ionization energy is

Physics
1 answer:
Marianna [84]3 years ago
6 0
The opposite of ionization energy is B, electronegativity. In short, we can define ionization energy as the amount of energy required to remove one electron from the outermost valence shell from an atom. Conversely, we can define electronegativity as an atom's "willingness" to accept one electron. 

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What happens if the breakdown voltage is exceeded.
s2008m [1.1K]

Answer:

this may help

Explanation:

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2 years ago
Light is shone on a diffraction grating
Pani-rosa [81]

Answer:

    λ = 482.05 nm

Explanation:

The diffraction phenomenon and the diffraction grating is described by the expression

         d sin θ = m λ

where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction

in this case they indicate the distance between slits, the angle and the order of diffraction

         λ = \frac{d sin \theta }{m}d sin θ / m

let's calculate

         λ = 1.00 10⁻⁶ sin 74.6 / 2

         λ = 4.82048 10⁻⁷ m

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         λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)

         λ = 482.05 nm

3 0
3 years ago
What was your train of thought as you navigated the picture of the candle?
defon

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Where is the picture

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4 0
3 years ago
Which of the following characteristics do all unicellular organisms share?
Fed [463]

Answer:

Asexual production they can be eukaryotes or prokaryotes

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8 0
3 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

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1 year ago
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