A sidereal day is the time it takes for
1 answer:
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Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
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● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
Answer:
g(x)=6(3)°x
Answer:
λ = 538.0 nm
Explanation:
The solution of the Schrödinger equation for the inner part of the well gives energy
= (h² / 8mL²) n²
Where n is an integer and L is the length of the well
They ask for the transition from the first excited state n = 2 to the base state n = 1
E₂ - E₁ = = (h² / 8mL²) (n₂² - n₁²)
Let's calculate
E₂-E₁ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ (0.7 10⁻⁹)²) (2² -1²)
E₂ –E₁ = 3.6968 10⁻¹⁹ J
Let's use the Planck equation
E = h f
c = λ f
E = h c / λ
E = E₂ ₂- E₁
h c / λ = 3.6968 10⁻¹⁹
λ = h c / (E₂-E₁)
λ = 6.63 10⁻³⁴ 3 10⁸ / 3.6968 10⁻¹⁹
λ = 5.380 10⁻⁷ m
Let's reduce
λ = 5.380 10⁻⁷ m (10 9 nm / 1 m)
λ = 538.0 nm