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nlexa [21]
2 years ago
13

Through how many volts of potential difference must an electron, initially at rest, be accelerated to achieve a wave length of 0

.27nm? A) 4.6 V
B) 0.046 V
C) 0.27 V
D) 4.6 MV
E)4.6 kV
Physics
1 answer:
deff fn [24]2 years ago
4 0

Answer:

20.7 volts

Explanation:

m = mass of electron = 9.1 x 10⁻³¹ kg

λ = wavelength of electron = 0.27 x 10⁻⁹ m

v = speed of electron

Using de-broglie's hypothesis

λ m v = h

(0.27 x 10⁻⁹) (9.1 x 10⁻³¹) v = 6.63 x 10⁻³⁴

v = 2.7 x 10⁶ m/s

ΔV = Potential difference through which electron is accelerated

q = charge on electron = 1.6 x 10⁻¹⁹ C

Using conservation of energy

(0.5) m v² = q ΔV

(0.5) (9.1 x 10⁻³¹) (2.7 x 10⁶)² = (1.6 x 10⁻¹⁹) ΔV

ΔV = 20.7 volts

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Explanation:

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The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

or h = \frac{4186x140}{69x10}  = 850 m

If the 20% of the total energy is used

the height h₀ = \frac{0.2x4186x140}{69x10} = 170 m

5 0
2 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
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Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

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Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
2 years ago
3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in
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2 years ago
A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the
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Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written asx(t)=Asin(\omega t+\phi)

'A' is the amplitude = 6 inches (given)

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At equilibrium we have

mg=kx\\\\k=\frac{mg}{x}

Applying values we get

k=40 lb/ft

thus natural frequency equals

\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}

Thus the equation of motion becomes

x(t)=6sin(8t+\phi)

At time t=0 since mass is at it's maximum position thus we have

A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})

Thus the position of mass at the given times is as follows

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2) at \frac{\pi}{8} x(t)=5.9909inches

3) at \frac{\pi}{6} x(t)=5.98397inches

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4 0
1 year ago
The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
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Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

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B=\frac{\mu _o I}{2\pi r}

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           r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

   

7 0
2 years ago
Read 2 more answers
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