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2 years ago

What is responsible for the production of x ray emission at the cutoff wavelength?

Physics

1 answer:

Katen [24]2 years ago

3 0

Answer:

correct answer is option 5 (The emission by the incident electron when it loses its energy in a single emission)

Explanation:

The emission by the incident electron when it loses its energy in a single emission then X-rays of minimum wavelength which is also known as X-rays of cut-off wavelength are emitted.

If potential V is applied on coolidge tube then cut-off wavelength of the continuous X-ray spectrum is related to the accelerating potential V by following equation

$\lambda_{cut-off} =\frac{hc}{eV} =\frac{12400}{V}$ Angstrom/V

You might be interested in

If a 2 kg spring is compressed by exerting a force of 10 newtons a distance of 50 cm how much work is done?

Elina [12.6K]

Convert cm into meters,
50/100=0.5m

work done= 10x0.5
=5J

we dont consider the weight of the spring as it acts downwards.

5 0

2 years ago

Adjacent antinodes of a standing wave on a string are 15.0 cmapart. A particle at an antinode oscillates in simpleharmonic motio

viva [34]

Answer:

a) D = 15 cm , b) λ = 30.0 cm , c) 0.0850 cm, d) v = 400 cm / s , e) v = 80.43 cm / s , v = -80.43 cm / s, f) D₂= 7.5 cm

Explanation:

Standing waves form when two waves of the same frequency travel in opposite directions,

A) in the waves the distance of the nodes and antinode are the same since the wavelength is constant

D = 15 cm

B) the wavelength is the distance for which the wave repeats itself, in the case of a standing wave, the distance between two nodes is lamita of the wavelength.

D = λ / 2

λ= 15 2

λ = 30.0 cm

C) the amplitude of each wave is 0.0850 cm, the amplitude of the standing wave is double A = 0.17 cm

D) Let's use the speed ratio

v = λ f

f = 1 / T

v = λ / T

v = 30.0 /0.0750

v = 400 cm / s

E) the transverse speeds are the speed of the oscillatory movement

y = A cos (wt)

w = 2π f = 2π / T

w = 2π / 0.0750

w = 83.78 rad / s

y = 0.850 cos (83.78 t)

Speed is

v = dy / dt

v = -A w cost wt

v = - 0.850 83.78 cos (83.78 t)

v = -80.43 cos (83.78 t)

The maximum speed when the cosine values ±1

v = 80.43 cm / s

v = -80.43 cm / s

F) if we draw a drawing, the distance between two nodes is half the wavelength, at the distance between an antinode synod is half this, it occupies a quarter of the wavelength

D₂ = ¼ λ

D₂ = 30.0/4

D₂= 7.5 cm

5 0

3 years ago

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

2 years ago

What rock type would most likely be found in the location marked by the black rectangle in the diagram below?

pychu [463]

There is no diagram below so I can't answer the question

5 0

2 years ago

Read 2 more answers

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

oksian1 [2.3K]

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

3 0

2 years ago

Read 2 more answers