I think it’s manufacturing
$2.
Both tickets cost $1.50
$1.50 x 2 = $3
$5 - $3 = $2
The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.
<h3>How does kVp impact the exposure to digital receptors?</h3>
The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.
<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>
Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.
To know more about kVp visit:-
brainly.com/question/17095191
#SPJ4
Answer:
Qx = 9.10
m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 ×
Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85
)²× 9
× sin18 × cos18
Qd = 94.305 ×
m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 ×
× π × 85
× ( 9
)³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 ×
m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 ×
- 85.2 ×
Qx = 9.10
m³/s