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4 years ago

When the speed of electrons striking the anode is increased, the of the ______ of the emitted x-rays increases

Physics

1 answer:

Kipish [7]4 years ago

8 0

The speed of electrons striking the anode is increased, the density of the emitted x-rays increases.

Answer: Option 2

<u>Explanation:</u>

Whenever electrons are made to be incident on any anode, they will be interacting with the surface atoms of anode. So these interaction will lead to excitation of electrons of the surface atoms of the anode.

Thus when the excited electron will come back to ground state, it will emit x-rays. Thus, on increasing the speed of the striking electrons, it will gain the property of interacting with larger amount of surface atoms leading to increase in the emission of X-rays .

Thus, this increase in the emission of X-rays will lead to increase in the density of X-ray. So the speed of electrons striking the anode is increased, the density of the emitted x-rays increases.

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How do you get a tachyionic particle and how are they faster than light

melomori [17]

<span>Tachyons are studied in an area called particle physics, and I must say this is a bit out of my league, but I'll give you some general thoughts. Tachyons are hypothetical particles resulting from what physicists call a thought experiment. Back in the 1960s, some physicists wondered what would happen if matter could travel faster than the speed of light, something that is supposed to be impossible according to the Theory of Relativity. So these particles may or may not exist because they have not been proven or disproven by real experiment as of yet. What people have done is apply existing formulas to the unique properties of tachyons (like imaginary mass!). What comes out is a particles that go faster when they lose energy with a MINIMUM velocity of the speed of light and a maximum velocity of infinity! Hope that helps Ben, theoretical physics is a weird place and is not too far off from philosophy.</span>

4 0

3 years ago

An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a con

antoniya [11.8K]

(a) 4.03 s

The initial angular velocity of the wheel is

$\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s$

The angular acceleration of the wheel is

$\alpha = -3.40 rad/s^2$

negative since it is a deceleration.

The angular acceleration can be also written as

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 0$ is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

$t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s$

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

$\theta= \omega_i t + \frac{1}{2}\alpha t^2$

where we have

$\omega_i=13.7 rad/s$ is the initial angular velocity

$\alpha = -3.40 rad/s^2$ is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

$\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad$

6 0

3 years ago

Before partacting, if correct marked as brainest

Molodets [167]

Answer:

C. Basic swimming capability.

Explanation:

one of the way before participating in any water-based sport is know how to do the basic swimming way.

7 0

3 years ago

A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 4

san4es73 [151]

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
$\Sigma F = T - F_g$

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
$F_c = \frac{mv^2}{r}$

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
$\frac{mv^2}{r} = T - Mg$