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Fiesta28 [93]
3 years ago
5

The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10

−18 JJ of kinetic energy in this distance?

Physics
2 answers:
Serjik [45]3 years ago
8 0

Answer:

a=1.1*10^{18}\frac{m}{s^2}

Explanation:

We use the following kinematic formula to calculate the acceleration:

v_f^2=v_0^2+2ax

The kinetic energy is defined as:

\Delta K=\frac{m(v_f^2-v_0^2)}{2}\\v_f^2-v_0^2=\frac{2\Delta K}{m}

Replacing this in the acceleration formula and solving for a:

\frac{2\Delta K}{m}=2ax\\a=\frac{\Delta K}{mx}\\a=\frac{2*10^{-18}J}{(9.1*10^{-31}kg)(2*10^{-6}m)}\\a=1.1*10^{18}\frac{m}{s^2}

Ilya [14]3 years ago
4 0

The solution is in the attachment

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Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?
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What are the difficulties with the capture hypothesis of the Moon’s origin?
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3 years ago
a freight train travels at v=60(1-e^-t) ft/s, where t is the elapsed time in seconds. Determine the distance traeled in tree sec
lesantik [10]

Answer

given,

v = 60(1-e^{-t})\ ft/s

t = 3 s

we know,

v = \dfrac{dx}{dt}

a = \dfrac{dv}{dt}

position of the particle

dx = v dt

integrating both side

\int dx =\int 60(1-e^{-t}) dt

x = 60 t + 60 e^{-t}

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Distance traveled by the particle in 3 s is equal to 182.98 ft

now, particle’s acceleration

a = \dfrac{dv}{dt}

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at t= 3 s

  a = 60 e^{-3}

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acceleration of the particle is equal to 2.98 ft/s²

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3 years ago
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