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2 years ago

What substance can be used to electrolyze water?

Physics

1 answer:

lubasha [3.4K]2 years ago

6 0

C/any electrolyte that is not easily reduced or oxidized

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What's the minimum Out PUT WORK<br> required to raise 14,0m3 of water 26.0m?

BartSMP [9]

Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

3 0

2 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

2 years ago

Read 2 more answers

A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

6 0

2 years ago

A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the

zlopas [31]

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

$a=2.5 m/s^2$

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

$d=0+\frac{1}{2}(2.5)(35)^2=1531 m$

8 0

2 years ago

How much time would it take for an object to fall 4.7 meters

levacccp [35]

Answer:

4.7 is 10 as much as the number 0.47.

If you multiply 0.47 x 10 it will equal 4.7

Explanation:

4 0

2 years ago