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Liula [17]
3 years ago
10

Let's assume you use green light (λ = 550 nm) to look at an electron. What is the uncertainty in determining the electron's velo

city? Express your answer rounded up to the nearest hundredth.
Physics
1 answer:
Rudiy273 years ago
8 0
By uncertainty principle

λ = h / p.         

Where λ = wavelength,  h = Planck's constant = 6.63 * 10⁻³⁴ Js

λ = wavelength = 550 nm = 550 * 10⁻⁹ m, Mass of electron = 9.1 * 10 ⁻³¹ kg

p = Momentum = mv

λ = h / mv

v = h / mλ

v = 6.63 * 10⁻³⁴ / (9.1 * 10⁻³¹ *  550 * 10⁻⁹)

v = 1 324 675.325 m/s

v ≈ 1.325 *10⁶  m/s
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Which of these is a characteristic of the Milky Way galaxy
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A

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The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C
Mashcka [7]

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

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8 0
3 years ago
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
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  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
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Vsevolod [243]

Answer:

The answer is for your question is :

Explanation:

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