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lozanna [386]
3 years ago
15

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

w many electrons are accelerated per pulse? (b) What is the average current for a machine operating at 500 pulses/s? If the electrons are accelerated to an energy of 50 MeV, what are the (c) average power and (d) peak power of the accelerator?

Physics
2 answers:
e-lub [12.9K]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

Crank3 years ago
5 0

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

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Which electromagnetic wave has the lowest frequencies (less than 3×109 hertz)?
ICE Princess25 [194]

Answer:

Radio waves

Explanation:

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Radio waves are the electromagnetic waves with lowest frequency, their frequency is lower than 300 GHz (3\cdot 10^{11} Hz) and therefore they are the electromagnetic waves with lowest energy (in fact, the energy of an electromagnetic wave is proportional to its frequency). They are generally used for radio and telecommunications since this type of waves can travel up to long distances.

5 0
3 years ago
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Standing still, Bruce, the quarterback, gets tackled by Biff, the 90.0-kg tackle, who is traveling at 7.0 m/s. Upon collision, B
tatyana61 [14]

\\ \sf\longmapsto \Delta P=P

\\ \sf\longmapsto m1v1=m2v2

\\ \sf\longmapsto 90(7)=m2(10)

\\ \sf\longmapsto 10m2=630

\\ \sf\longmapsto m2=\dfrac{630}{10}

\\ \sf\longmapsto m2=63kg

Bruces mass is 63kg

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3 years ago
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LOTS OF BRAINLIST WILL BE GIVING TO THOSE WHO HELP
Alex_Xolod [135]
You know you can skip those and just submit them, they don’t even check them
4 0
3 years ago
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Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of an average person i
faust18 [17]

Well first of all, you must realize that it depends on how the jumpers are distributed on the earth's surface.  If,say, one billion of them are in the eastern  hemisphere and the other billion are in the western one, then the sum of all of their momenta could easily be zero, and have no effect at all on the planet.  I'm pretty sure what you must have in mind is to consider the Earth to be a block, with a flat upper surface, and all the people jump in the same direction.

average mass per person = 60 kg.
jump velocity = 7 m/s straight up and away from the block, all in the same direction
one person's worth of momentum = (m) (v) = 420 kg.m/s
sum of two billion of them = 8.4 x 10¹¹ kg-m/s all in the same direction

Earth's "recoil" momentum = 8.4 x 10¹¹ in the opposite direction = (m) (v)

Divide each side by 'm' :     v = (momentum) / (mass) =

The Earth's "recoil" velocity is   (8.4 x 10¹¹) / (5.98 x 10²⁴) = 

                                                               1.405 x 10⁻¹³ m/s =

                                              <em> 0.00000000014 millimeter per second

</em>
I have no intuitive feeling for this kind of thing, so can't judge whether
the answer is reasonable.  But my math and physics felt OK on the
way to the solution, so that's my answer and I'm sticking to it.

4 0
3 years ago
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