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.
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Answer: b) FCOR( Face-centered orthorhombic)
Explanation: Face centered orthorhombic lattice is the lattice that has eight lattice corner points and they also have each face with center lattice point.This lattice structure is usually not common in metals.Whereas the face centered lattice structure , body structure lattice and hexagonal close packed lattice are common in metal.Thus option (b) is the correct option.
Answer:
a) 15.37 mm
b)
c) 5.7186 W/m². K
d) 0.60 m
Explanation:
Given that :
The surface temperature = 130°C = ( 130+ 273 ) K = 473 K
suspended in quiescent air at 25°C = ( 25 + 273 ) K = 298 K
Atmospheric Pressure = 1 atm
The properties obtained from Table A - 4 include :
v =
k = 0.03 W/m K
Pr = 0.700
η = 5
Hence, the boundary layer thickness at a location 0.15 m measured from the lower edge is 15.37 mm
b) The maximum velocity in the boundary layer with f'(n) = 0.275
u = 0.3659 m/s
the maximum velocity in the boundary layer at this location is 0.3659 m/s
the position in the boundary layer where the maximum occur is calculated as:
3.074 mm
c) Using the similarity solution result, , determine the heat transfer coefficient 0.15 m from the lower edge.
we know that:
=
= 28.593
Making the subject from the above formula:
= 5.7186 W/m². K
d) to determine the location on the plate that the boundary layer we become turbulent ; we have the following:
0.60 m