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Kay [80]
3 years ago
9

What is the wavelength associated with an electron with a velocity of 4.8X10s m/s? (Mass of the electron is 9.1X10-31kg)?

Physics
1 answer:
amid [387]3 years ago
4 0

Answer:

1.52 nm

Explanation:

Using the De Broglie wavelength equation,

λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s

So, λ = h/p

λ = h/mv

substituting the values of the variables into the equation, we have

λ = h/mv

λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)

λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)

λ = 0.1518 × 10⁻⁸ m

λ = 1.518 × 10⁻⁹ m

λ = 1.518 nm

λ ≅ 1.52 nm

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A 18.4-kg box rests on a frictionless ramp with a 16.1° slope. The mover pulls on a rope attached to the box to pull it up the i
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Answer:

F= 56,1 N :

Explanation:

We apply Newton's first law for balancing forces system.

We take the x axis in the direction of the ramp with a 16.1° slope.

∑Fx= 0

∑Fx: algebraic sum of forces ( + to the right, - to the left)  

Problem development:

Look at the free body diagram :

The only forces that act on the box are the weight and tension of the rope because there is no friction.

T: Rope tension (N)

W :Box weight (N)

∑Fx= 0

Tx-Wx=0

Tx =Tcos( 43.2°- 16.1°)= Tcos ( 27.1°)

Wx= W*sen  16.1°= m*g*sen  16.1°=18.4*9.8* sen  16.1° = 50N

Tcos ( 27.1°)-50=0

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T = (50) / (cos ( 27.1°))

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What types of EM waves have you seen or used today? Describe them.
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3 years ago
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PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!
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Answer:D

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4 years ago
An electric motor has an effective resistance of 22 22 and an inductive reactance of 72 12 when working under load. The voltage
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Answer:

Current amplitude, I = 5.57 A

Explanation:

It is given that,

Effective resistance, R = 22 ohms

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The total impedance of the RL circuit is given by :

Z=\sqrt{R^2+L^2}

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From Ohm's law,

V=I\times Z

I=\dfrac{V}{Z}

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I = 5.57 A

So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.

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