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Growth and development. Hospitals. Medicine of catastrophe. Examinations exercises. Solution;. ( h - prophylactic dose (one biodose =5000 MER)
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Growth and development
58. The main signs of growth and development of children. The methods of assessment. Acceleration.
59. Prophylactics of the skeleton deformation of children.
60. Prophylactics respiratory diseases of children.
61. Prophylactics of cardiovascular diseases of children.
62. Prophylactics of gastrointestinal diseases of children.
63. Prophylactics of myopia of children.
64. Hygiene of school lessons.
65. Exhausting of school children, the aetiology, preventive methods of exhausting.
66. Medical supervision of the sports in the school.
67. Hygienic recommendation of kinder-garden planing.
68. Organisation of daily routine in kinder-gardens.
69. Hygienic recommendation of school planing.
Hospitals
70. Hygienic evaluation of hospital planing systems, and the locality in the city.
71. The hygienic needs for planing the department of internal diseases.
72. The hygienic needs for planing the department of surgical and gynaecological diseases.
73. The hygienic needs for planing the department of infectious diseases.
Medicine of catastrophe
74. Medicine of catastrophe. The main problems due to catastrophe.
75. The externals factors in relation with results of catastrophe.
76. The external factors and their influence on the result of catastrophe, on the health of population.
77. Classification of catastrophe and its main problems.
78. The catastrophe in the chemical industries. The toxic effects of pollution on the health of populations.
79. The role of doctors in resolving the consequence of catastrophe.
80. The sanitary and hygienic measures to resolve the result of catastrophe.
81. The principles of resolving the housing problem. Purification of drinking water. Nutrition in the area of catastrophe.
82. Role of doctors in resolving the sanitary problems as a results of catastrophe.
EXAMINATIONS EXERCISES
1. Write your conclusion and recommendation about microclimate in the class. Temperature 25°c, alternative humidity 80%, the air velocity 0, 1 m/s.
Solution;
a) The normal level of temperature in class is 18-20°
b) The normal level of humidity is 40-60%
c) The normal level of air velocity is 0, 2-0, 4 m/sec
That is mean, the temperature and the humidity higher than
the normal level and the air velocity is lower than normal level.
The class needs for artificial and natural ventilation, because
the microclimate effects the thermoregulation of the students body.
2. The light coefficient (Ic - ck) of the class on the ground flour is 1: 5, coefficient of the natural light (cnl - keo) on the last desk: 1, 1%. Your conclusion about the amount of the light in that class.
Solution: In the class the normal level of CK is 1: 4 - 1: 6. The level of KEO is not less than 1, 5%.
Conclusion: The amount of natural light in the class is normal. But the amount of the light on the last desk is lower than the normal level.
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Recommendation: We need to clean the windows and the colour of the wall must be having light colour.
3. Find out the number of lamps - EUF-15, in class which are used as prophylactic lamps for schoolchildren's. The daily dose equal 0, 5 biodose, area of the class 48m2, the exposure time 4 h.
Solution: The formulas for finding the number of lamps are:
5, 4 x S x H
F = —————
T
S - area of the room
( H - prophylactic dose (one biodose =5000 MER)
EUF15(F1) - 340 MER
T - exposure time in mounts
5, 4 x 48 x 2500 2698
F = -——————— = 2698wer, n = ——----- = 7, 8(8) lamps
240 340
4. The playing room in the kinder-garden, its area 50m2, the height of the ceiling is 3m, it need for air disinfection. How many bacte-riocide lamps BUF-15 you need for disinection of that room. Show the location of the lamps, and the time of disinfection, in the presence of the children.
Solution: For disinfection of one m3 of the air we need one wt. The capacity of the room is 3-50=150 m\ that means we need 150 wt, the power of one lamp BUF - 15 = 15 wt.
The number of the lamps = 150: 15 = 10 lamps. If children are present in the room we need a metallic screen on the lamp.
The location of the lamps near about the door and the windows. The time of disinfection not more than 2h, after that we must ventilate the room.
5. For prophylactic purposes for 40 healthy children, you have to use lamp PRK-7. The area of the hall = 180m2. Find out the number of children expossuring at one time, select the schedule of expossuring for 10 days. Solution:
To receive one biodose the distance from the lamp must be 3m, time of exposuring 3, 7 min (If the distance 2m the time must be 1, 6 min, if the distance 1m the time must be 0, 5 min).
To find out the number of children's exposuring at one time, use the formula 2pR. 2. 3, 14. 3 = 18, 8, the distance between children's 0, 8- 1m, that mean at one time we can exposure 20 children's. The schedule for exposuring
| day | ||||||||||
| biodose | 0, 5 | 0, 5 | 1, 5 | 2, 5 | 3, 5 | 4, 5 |
6. The distance from radioactive iodine is 2m, its physical dose equal to 45 mkv/sec. Find out the physical dosage of this element if the distance 3m?
Solution:
You have to use the formula:
D1 r 2 ²
F = —— = ——
D2 r 1²
D - dose r – distance
7. After working 36 hours with radioactive element the exosuring dose of laboratory assistant is 03 roentgen. Your conclusion and recommendation about the condition on the work.
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Solution:
The PDD for week (36 hours) must be 0, 1 ber
I ber =l rad =1, 14 roentgen
That mean 0, 3 roentgen 0, 342 ber.
He received 3 times more than the normal (PDD 0, 1)
Recommendation: he cannot work till 2, 5 weeks
8. Find out exposuring dose of doctor who worked with gamma iozining source, if the activity of that source is 5mg. eq. rad. He is working every day for 6h. The distance from the source is 40 cm.
Solution : To find out the dose you have to used the formula:
8, 4 x m x t
D = —————
r2
D - dose m - activity t - time r - distance
9. Find out the distance from the radioactive source where the nurse is working for 6 h daily. The activity of the gamma source is 20 mg/equl/rad. Solution:
use the formula: m x t
120 = —————
r2
m - activity t - time for one week r – distance
10. Find out the thickness of screen to reduce the intensity of gamma iozining for 100 times, if the thickens of lead 1, 8 cm it will be reduced 50% of the intensity.
Solution:
To reduce the activity 4 times you have to use 2 layers, 8 times = 3 layers, 16 times = 4 layers, 32 times = 5 layers, 64 times = 6 layers, for 128 times = 7 layers.
That means, to reduce the activity of gamma iozinig for
100 times = 7 • 1, 8 = 12, 6 cm the screen thickness (use special table).
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