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Artyom0805 [142]
3 years ago
12

An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 106 V/m. (a) What energy in keV is give

n to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
Physics
2 answers:
vladimir1956 [14]3 years ago
7 0

Answer:

a) 800 keV

b)  24.996 km.

Explanation:

(a) we have

\large \Delta K.E=q\Delta V  .............(1)

where,

\large \Delta K.E = Change in kinetic energy

q = charge of an electron

\Delta V = Potential difference

also

\large E=\frac{V}{d}       .......(2)

E = electric field

d = distance traveled

Now from (1) and (2) we have,

\large \Delta K. E=qV=qEd

substituting the values in the above equation, we get

\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})

\large \Delta K. E=800keV

Thus, the energy gained by the electron is <u>800 keV</u> if it is accelerated over a distance of 0.400 m.

(b) Using the equation (1), we have

\large d=\frac{\Delta K.E}{qE}

\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})

or

\large d=2.4996\times 10^4m

or

\large d=24.996\times 10^3m=24.996km

Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of<u> 24.996 km.</u>

ohaa [14]3 years ago
5 0

Answer:

a) 800 keV

b) 25 km

Explanation:

Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\

b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km

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Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

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