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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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Explanation:

Given that,

Position of the particle at t = 0,

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Velocity of the particle at t = 0

u=(1i+6j)\ m

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a=(5i+7j)\ m/s^2

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

v=u+at

v=(1i+6j)+(5i+7j)10

v=(51i+76j)\ m/s

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

y'=y+ut+\dfrac{1}{2}at^2

y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2

y'=\dfrac{19}{2}i+\dfrac{39}{2}j

(c) Magnitude of y',

|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}

|y'| = 21.69 meters

Direction of the y',

tan\theta=\dfrac{y}{x}

tan\theta=\dfrac{39/2}{19/2}

\theta=64.02^{\circ}

Hence, this is the required solution.

4 0
2 years ago
On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som
k0ka [10]

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

v=f \lambda

Where:

v is the speed

f=\frac{1}{T} is the frequency, which has an inverse relation with the period T=1 h

\lambda=800 km is the wavelength

Solving with the given units:

v=\frac{1}{T}\lambda

v=\frac{1}{1 h}800 km

v=800.00 km/h This is the speed of the wave in km/h

Transforming this speed to m/s:

v=800.00 \frac{km}{h} \frac{1 h}{3600 s} \frac{1000 m}{1 km}

v=222.22 m/s This is the speed of the wave in m/s

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3 years ago
What is the kinetic energy of a 14kg object traveling at 10m/s
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Answer:

Explanation:

kinetic energy=1/2*mass*velocity^2

=1/2*14kg*10^2

=7*100

=700 joule

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2 years ago
What is the refractive index of air if light travels through it at 3.0 108 m/s?
olchik [2.2K]

Answer:

n = c/v = (3.00 x 108 m/s)/(2.76 x 108 m/s) = 1.09. This does not equal any of the indices of refraction listed in the table.

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He user of a machine applies force to the machine over the _____ distance.
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The blank distance is your answer

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