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nlexa [21]
1 year 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]1 year 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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Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

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8 0
1 year ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

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6 0
2 years ago
A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.
sladkih [1.3K]

Answer:

Explanation:

Potential due to a charged metallic sphere having charge Q and radius r on its surface will be

v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre  potential is

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a ) On the surface of the shell , potential due to positive charge is

V₁ = \frac{9\times10^9\times5\times10^{-6}}{6\times10^{-2}}

On the surface of the shell , potential due to negative  charge is

V₁ = \frac{- 9\times10^9\times5\times10^{-6}}{6\times10^{-2}}

Total potential will be zero . they will cancel each other.

b ) On the surface of the sphere potential

= \frac{9\times10^9\times5\times10^{-6}}{2\times10^{-2}}

= 22.5 x 10⁵ V

On the surface of the sphere potential due to outer shell

= \frac{9\times10^9\times5\times10^{-6}}{5\times10^{-2}}

= -9 x 10⁵

Total potential

=( 22.5 - 9 ) x 10⁵

= 13.5 x 10⁵ V

c ) In the space between the two , potential will depend upon the distance of the point from the common centre .

d ) Inside the sphere , potential will be same as that on the surface that is

13.5 x 10⁵ V.

e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.

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It will be <span>Einstein's General Relativity</span>
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The two bols of ice creams are composed of vanilla the same subtances
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