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Norma-Jean [14]
3 years ago
12

An electron and a proton are separated by a distance of 1.0 m. What happens to the force between them if the electron moves 0.5

m away from the proton?
Physics
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Explanation:

It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them

F=k\dfrac{q_1q_2}{1^2}...............(1)

We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.

F'=k\dfrac{q_1q_2}{0.5^2}...............(2)

On dividing equation (1) and (2) we get :

\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}

\dfrac{F}{F'}=0.25

F' = 4 F

So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.

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azamat

Answer:

a) wavelength = 656.3 nm

b)  the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

Explanation:

Given that;

angle of diffraction Θₓ = 22.78°

incident angle Θ₁ = 0

slit separation d  = 5900 lines per cm = 1/5900 cm = 10⁻²/5900 m = 0.01/5900 m

order of diffraction n = 1

wavelength λ = ?

to find the wavelength, we use the expression

λ = d (sinΘ₁ + sinΘₓ) / n

To find the wavelength λ;

λ = 0.01/5900 × (sin0 + sin22.78° )

λ = 6.5626 × 10⁻⁷ m

λ = 656.3 x 10⁻⁹ m

∴ λ = 656.3 nm

b)

According Balnur's  series spectral lines; n₁ = 3, n₂ = 2 and

λ = R [ 1/n₂² - 1/n₁²]

where  R is Rydberg's constant

from λ = R [ 1/n₂² - 1/n₁²]

R = 1/λ [n₂²n₁² / n₁² - n₂²]

R = 10⁹/ 656.3 [ 9 × 4 / 9 - 4 ]

R = 1.097 × 10⁷ m⁻¹

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3 years ago
A coin has a radius of 1.06 cm and a thickness of 1.2 mm. Find its<br><br> volume in m^3.
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Answer:

The volume of the coin is 4.236 x 10⁻⁷ m³

Explanation:

Given;

radius of a coin, r = 1.06 cm = 0.0106 m

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Area of the coin = πr² = π (0.0106)² = 3.5304 x 10⁻⁴ m²

The volume of the coin = (3.5304 x 10⁻⁴ m²) x (0.0012 m)

The volume of the coin = 4.236 x 10⁻⁷ m³

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Answer:

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