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aleksandrvk [35]
2 years ago
14

Two metal plates 4 cm long are held horizontally 3 cm apart in a vacuum, one being vertically above other. The upper plate is at

a potential of 3000 & the lower is earthed. Electrons having a velocity of 1×10^7 m/s are injected horizontally midway between the plates & in the direction parallel to 4 cm edge. Calculate the vertical deflection of the electron beams as it emerges from the plates. (Assume “e/m=1.8×10^11 c/kg”).
Physics
1 answer:
kati45 [8]2 years ago
5 0

Answer:

the answer to your question is 4 cm long

Explanation:

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love history [14]

Answer:

The correct answer is 24.

Explanation:

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3 years ago
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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
3 years ago
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Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

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Where is most of the mass of an atom located? A.in the nucleus B.in the orbits C.in the electrons Dit is split between the nucle
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