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Misha Larkins [42]
2 years ago
10

Horizontal angulation is: Select one: a. the side-to-side angulation. b. different when using the paralleling and bisecting tech

niques. c. correct when the central ray is parallel to the curvature of the arch. d. determined by the Stabe bite-block.
Physics
1 answer:
statuscvo [17]2 years ago
8 0

Horizontal angulation is Select one: a. the side-to-side angulation. b. different when using the paralleling and bisecting techniques.

Angles that are horizontal. relates to the central ray's placement in a horizontal, or side-to-side, plane.

A picture with an overlap of nearby structures in the horizontal plane is produced by situating the central ray such that the horizontal angulation is not directed through the interproximal contacts of the adjacent teeth (the contact areas of the teeth are superimposed over each other).

Fracture angulation refers to a particular kind of fracture displacement in which the bone's natural axis has been changed so that the distal end now points off in a different direction. When using a bite-wing tab, the x-ray beam's center ray must be pointed at the contact points between teeth. The x-ray beam must be centered on the receptor to guarantee that the receptor is exposed.

Learn more about horizontal angulation  here brainly.com/question/28043105

#SPJ4.

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The potential energy of the car when it let go is 20,000 J.

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The given parameters;

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<em />

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the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

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F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

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e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

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