Psoas major muscles, kidneys, inferior ribs, and lumbar transverse processes are demonstrated on AP abdominal projections.
<h3>What are AP abdominal projections?</h3>
The anteroposterior (AP) radiograph the patient in a supine position is the basis of the majority of plain-film examinations of the abdomen.
However, an upright-position film could also be helpful in patients with suspected bowel obstruction, also as to assess the air-fluid levels in the distended bowel.
This information also can be obtained on a film of the abdomen taken with the patient in the lateral decubitus position.
An AP supine projection is usually called a flat plate or KUB (because it includes the kidneys, ureter, and bladder)
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Answer: The molar mass of is 67.81 g/mol
Explanation:
Molar mass is defined as the mass in grams of 1 mole of a substance.
S.I Unit of Molar mass is gram per mole and it is represented as g/mol.
Atomic Mass of Boron (B) = 10.81 g
Atomic Mass of Flourine (F) = 18.99 g
Molecular mass of = 1 (10.81)+3(18.99) g = 67.81 g
Thus molar mass of is 67.81 g/mol
Carbon- C
hydrogen - H
Nitrogen - N
Iron- Fe
Copper- Cu
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
The answer is; D
While electromagnetic waves travel at an astounding 3.00 x 10^8 m/s and any delay seems imperceptible in short distances, in long distances, the waves take some time to reach the other end. This is why there is usually a small delay when ground communication tries to reach astronauts in space. Also, remember that when a communication is relayed from the earth, it has to reach the destination and then wait for a response back to earth which covers the same distance or longer/shorter if the target is moving.
The distance to the moon is 384,400 km, therefore multiply this by 2 = 768,800 km
768,800,000m/300,000,000m = 2. 56 seconds
Therefore radio waves sent to the moon from earth will have a minimum 2.5 seconds delay not considering the processing time of this communication by the destination before sending feedback. ‘Over to you’ signals end of a message by the messenger hence allowing the other messenger on the other to respond.