College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)
3rd Edition
ISBN: 9780134201979
Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher: PEARSON
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Chapter P.6, Problem 8P
The following passages and associated questions are based on the material of Part VI
Taking an X Ray
Cathode Turret electrode
X rays are a very penetrating form of
FIGURE VI.1
A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, treeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest.
An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can't be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA · s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted.
Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problem that follow.
What, physically, does the product of a current (in mA) and a time (in s) represent?
- A. Energy in mJ
- B. Potential difference in mV
- C. Charge in mC
- D. Resistance in mΩ
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In the figure what is the net electric potential at point P due to the four particles if V = 0 at infinity, q = 2.12 fC, and d = 1.75 cm?
d
Number
MI
Units
+q
Current Attempt in Progress
In the figure what is the net electric potential at point P due to the four particles if V = 0 at infinity, q = 2.12 fC, and d = 1.75 cm?
d
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MI
Units
+q
A 0.500 kg sphere moving with a velocity given by (2.00î – 2.60ĵ + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with an initial velocity of (−1.00î + 2.00ĵ – 3.20k) m/s.
(a) The velocity of the 0.500 kg sphere after the collision is (-0.90î + 3.00ĵ − 8.00k) m/s. Find the final velocity of the 1.50 kg sphere.
R =
m/s
Identify the kind of collision (elastic, inelastic, or perfectly inelastic).
○ elastic
O inelastic
O perfectly inelastic
(b) Now assume the velocity of the 0.500 kg sphere after the collision is (-0.250 + 0.850ĵ - 2.15k) m/s. Find the final velocity of the 1.50 kg sphere.
✓ =
m/s
Identify the kind of collision.
O elastic
O inelastic
O perfectly inelastic
(c) Take the velocity of the 0.500 kg sphere after the collision as (−1.00ỉ + 3.40] + ak) m/s. Find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a positive value and a negative value. Report each with their
corresponding final velocities.)
a…
Chapter P Solutions
College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)
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