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Fed [463]
3 years ago
5

In the diagnostic radiology energy range (which includes mammography) from 23 to 150 kVp, which of the following tissues possess

es the greatest ability to absorb radiant energy through the process of photoelectric absorption?
Physics
1 answer:
Semmy [17]3 years ago
4 0

Answer:

Bone

Explanation:

Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.

The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.

The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.

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5 0
3 years ago
1. A 180 kg motorcycle travels in a straight line on a horizontal road. The relationship between
Sergio039 [100]
Eh not really sure bout this one
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3 years ago
When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi
ryzh [129]

Answer:

<u>The pendulum bob swing past the mean position because:</u>

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to  the restoring toque the bob starts to accelerates towards its mean postion. The  maximum acceleration of the pendulum's bob  is -w^{2} Aand the the acceleration decreases as -w^{2} x  towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0
3 years ago
An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

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

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

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

If we re-arrange it for t, we get:

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

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a
jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A V_a=382\ V

Potential at V_c=785\ V

when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1

Change in Kinetic Energy of particle moving under the Potential From C to B

q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2

Divide 1 and 2 we get

\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}

on solving we get

V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c

V_b=\frac{743}{3}=247.67\ V

                     

4 0
3 years ago
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