Answer:
gravitational potential energy
Answer:
it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
Explanation:
The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.
To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates
The letter “j” is never found on the periodic table. As for numbers, there’s an infinite amount
The wavelength in nanometers of light when the energy is 1. 91 × 10^6 j for a mole of photons is <u>62. 8 nm.</u>
Wavelength is the distance among the same points (adjacent crests) within the adjoining cycles of a waveform signal propagated in space or along a cord. In wi-fi structures, this period is typically specified in meters (m), centimeters (cm), or millimeters (mm).
The wavelength is the distance between wave crests, and it is going to be the same for troughs. The frequency is the variety of vibrations that skip over a given spot in one 2nd, and it's far measured in cycles consistent with the second (Hz) (Hertz).
Frequency is the ratio of pace and wavelength in relation to speed. In comparison, wavelength refers to the ratio of pace and frequency. Audible sound waves are characterized by way of a frequency range of 20 to 20 kHz. In contrast, the variety of wavelengths of visible light is from four hundred to seven hundred nm.
<u>calculation:-</u>
*E=hc/λ
1.91 × 10^6 J = (6.62610⁻³⁴) (3.00*10⁸) / λ
λ= (6.62610⁻³⁴) (3.00*10⁸) / 1.91 × 10⁶ J
λ= 1.0410⁻³¹× 10⁻⁹ × 6.022*10²³
=<u> 62. 8 nm </u>
Learn more about wavelength here:-brainly.com/question/10728818
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Answer:
x = 2,864 m
, Ra = 32.1 m
Explanation:
Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer
observer A β = 64 db
β = 10 log Iₐ / I₀
where I₀ = 1 10⁻¹² W / m²
Iₐ = I₀ 10 (β/ 10)
let's calculate
Iₐ = 1 10⁻¹² (64/10)
Iₐ = 2.51 10⁻⁶ W / m²
Observer B β = 85 db
I_b = 1 10-12 10 (85/10)
I_b = 3.16 10⁻⁴ W / m²
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
P = Ia Aa = Ib Ab
the area of a sphere is
A = 4π R²
we substitute
Ia 4pi Ra2 = Ib 4pi Rb2
Ia Ra2 = Ib Rb2
Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute
Ia (35 -x) 2 = Ib x2
we develop and solve
35-x = Ra (Ib / Ia) x
35 = [Ra (Ib / Ia) +1] x
x (11.22 +1) = 35
x = 35 / 12.22
x = 2,864 m
This is the distance of observer B
The distance from observer A
Ra = 35 - x
Ra = 35 - 2,864
Ra = 32.1 m
