The applications, either for diagnostic purposes or for therapeutic purposes, involve the use of X-rays----CT scan radiography ,external beam radiation therapy, fluoroscopy.
How is CT used for treatment planning?
CT planning enables more accurate localisation of both tumour and normal organs in addition to providing an accurate body contour and inhomogeneity corrections.
What is difference between CT scan and fluoroscopy?
Overall, fluoroscopy is a safe procedure, but potential risks include burns or radiation-induced injuries to the skin. On the other hand, CT scans are still snapshots of a “slice” of the body. They use X-rays to help your doctor view important organs
What is a fluoroscopic procedure?
During a fluoroscopy procedure, an X-ray beam is passed through the body. The image is transmitted to a monitor so the movement of a body part or of an instrument or contrast agent (“X-ray dye”) through the body can be seen in detail
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Answer:
6.21 rad/s
1.3041 m/s, 0.567 m/s²

Explanation:
= Final angular velocity
= Initial angular velocity = 0
= Angular acceleration = 2.3 rad/s²
= Angle of rotation
t = Time taken = 2.3 s
Equation of rotational motion

The angular speed is 6.21 rad/s
Linear velocity is given by

Linear velocity is 1.3041 m/s
Tangential acceleration is given by

Tangential acceleration is 0.567 m/s²

In degress the angle would be

From x axis it would be

The angle is
from x axis
Answer:
a = 1,008 10⁻³ m / s²
Explanation:
For this exercise, let's use the kinematic relations of accelerated motion
v² = v₀² - 2 a x
The negative sign is because the acceleration is opposite to the speed, the final speed is zero
0 = v₀² - 2 a x
a = v₀² / 2x
Let's reduce the magnitudes to the SI system
x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m
Let's calculate
a = 2.2²/2 2.4 10⁻³
a = 1,008 10⁻³ m / s²
Answer: 22 batches.
Explanation:
Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.
How many batches of cookies can you make with one bag of flour
Let's first convert 11 kg into grams (g) by multiplying it by 1000
11 × 1000 = 11000 g
Divide 11000 by 500
11000/500 = 22
Therefore, 22 batches of cookies can be made with one bag of flour.