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mestny [16]
3 years ago
5

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot

ons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.90 keV and a current of 4.95 mA produced by the generator. (a) What is the speed of the protons (in m/s)?
Physics
1 answer:
kari74 [83]3 years ago
8 0

Answer:

603383.67253 m/s

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

K = Kinetic energy = 1.9 keV

1\ ev=1.6\times 10^{-19}\ J

Velocity of proton is given by

v=\sqrt{\frac{2K}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 1.9\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=603383.67253\ m/s

The speed of the protons is 603383.67253 m/s

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castortr0y [4]
The potential energy of the block is given by:
V = m*g*h
m mass
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h height

The potential energy of a spring is given by:
V = 0.5 * k * x²

k spring constant
x compression of the spring

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3 years ago
A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate
yaroslaw [1]

Answer:

The  maximum error is  \Delta  \eta  = 2032.9

Explanation:

From the question we are told that

     The length  is  l  =  1\ m

      The radius is  r =  0.002 \pm  0.0002 \ m

        The pressure is  P  =  4 *10^{5} \ \pm 1750

        The  rate  is  v =  0.5*10^{-9} \ m^3 /t

       The viscosity is  \eta  =  \frac{\pi}{8} * \frac{P *  r^4}{v}

The error in the viscosity is mathematically represented  as

       \Delta  \eta  = | \frac{\delta \eta}{\delta P}| *  \Delta  P   +    |\frac{\delta \eta}{\delta r} |*  \Delta  r +  |\frac{\delta \eta}{\delta v} |*  \Delta  v

   Where  \frac{\delta \eta }{\delta P} =  \frac{\pi}{8} *  \frac{r^4}{v}

and         \frac{\delta \eta }{\delta r} =  \frac{\pi}{8} *  \frac{4* Pr^3}{v}

and          \frac{\delta \eta }{\delta v} =  - \frac{\pi}{8} *  \frac{Pr^4}{v^2}

So  

             \Delta  \eta  = \frac{\pi}{8} [ |\frac{r^4}{v}  | *  \Delta  P   +    | \frac{4 *  P * r^3}{v}  |*  \Delta  r +  |-\frac{P* r^4}{v^2}  |*  \Delta  v]

substituting values

            \Delta  \eta  = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}}  | *  1750   +    | \frac{4 *  4 *10^{5} * (0.002)^3}{0.5*10^{-9}}  |*  0.0002 +  |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2}  |*  0 ]

  \Delta  \eta  = \frac{\pi}{8} [56  +  5120 ]

   \Delta  \eta  = 647 \pi

    \Delta  \eta  = 2032.9

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