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2 years ago

Explain the mode of operation of x-ray

Physics

2 answers:

Serga [27]2 years ago

7 0

Answer:

If x-rays get in touch with our body tissues, they create a picture of a metal film. The high energizing rays are unable to penetrate soft tissue, such as skin and organs and travel through the laser. The places where the x-rays pass soft tissues are Black areas of an X-ray.

igomit [66]2 years ago

7 0

Answer:

Explanation:

Mode of Operation: In the X-ray tube a Low voltage high p.d is applied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. ... They are produced by increasing the p.d. between the cathode and anode.

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In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

7 0

3 years ago

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

3 years ago

Which of the following are correct statements about the way an atom is put

Ann [662]

Answer:

valenc e shell

Explanation:

4 0

3 years ago

According to the theory of plate tectonics

Darya [45]

Answer:

Alfred Wegener proposed the idea that earths lithospher is divided into tectonic plates.The theory was only accepted after he died.

8 0

3 years ago

Please help me whit this question

mariarad [96]

Answer:

Gypsum

Explanation:

Gypsum is a soft sulfate mineral composed of Calcium Sulfate di-hydrate. It is widely used in composition of fertilizer. It is also known as land plaster. On adding gypsum to soil, the quality of soil is improved. It conditions the soil and adds nutrients. The properties of soil is improved. Hence, the correct answer is Gypsum.

Hope this helped:)

pls mark brainlist

6 0

3 years ago

Explain the mode of operation of x-ray​

Explain the mode of operation of x-ray