Poziom Operacyjny - Nawigacja

1. Specjalista który przeprowadza kompensację kompasu magnetycznego i wyznacza tabelę dewiacji to:

dewiant

dewiator

kompensator

The correct answer is "dewiator". In Polish, "dewiator" refers to a specialist who conducts magnetic compass compensation and determines the deviation table. This person is responsible for adjusting the compass to ensure accurate navigation.

Explanation

The correct answer is "dewiator". In Polish, "dewiator" refers to a specialist who conducts magnetic compass compensation and determines the deviation table. This person is responsible for adjusting the compass to ensure accurate navigation.

2. Otrzymując z obliczeń różnicę długości >180 musimy:

dopełnić wyliczoną wartość do 360 i zmienić znak na przeciwny

dopełnić wyliczoną wartość do 180 i zmienić znak na przeciwny

dopełnić wyliczoną wartość do 360, a znak pozostawić bez zmian

When receiving a difference in length greater than 180, we need to complement the calculated value to 360 and change the sign to the opposite. This means that we need to add the calculated value to 360 and then change the positive value to negative or vice versa.

Explanation

When receiving a difference in length greater than 180, we need to complement the calculated value to 360 and change the sign to the opposite. This means that we need to add the calculated value to 360 and then change the positive value to negative or vice versa.

3. Statek A płynąc kursem rzeczywistym KR = 305 widział statek B dokładnie przed swoim dziobem. Statek B widział statek A na swoim prawym trawersie. Jakim kursem rzeczywistym płynął statek B ?

035 ̊

022 ̊

125 ̊

Statek A widział statek B przed swoim dziobem, co oznacza, że statek B znajdował się na kursie prostopadłym do kursu stateka A. Z drugiej strony, statek B widział statek A na swoim prawym trawersie, co oznacza, że statek A znajdował się na kursie równoległym do kursu stateka B. Jedynym kursem, który spełnia oba te warunki, jest kurs 035°.

Explanation

Statek A widział statek B przed swoim dziobem, co oznacza, że statek B znajdował się na kursie prostopadłym do kursu stateka A. Z drugiej strony, statek B widział statek A na swoim prawym trawersie, co oznacza, że statek A znajdował się na kursie równoległym do kursu stateka B. Jedynym kursem, który spełnia oba te warunki, jest kurs 035°.

4. Statek A znajdował się na pozycji 39.15'S, 73.12'W i płynął KD d = KR = 180 z prędkością nad dnem V d = 15 węzłów. Statek B znajdował się na pozycji 42.00'S, 177.55'E i płynął KD d = KR = 000 z prędkością nad dnem V d = 18 węzłów. Po ilu godzinach żeglugi statki zbliżą się do siebie na najmniejszą odległość?

4,6 godz.

5 godz.

7,2 godz

The two ships are moving towards each other, with ship A heading at a bearing of 180 degrees and ship B heading at a bearing of 0 degrees. The difference in longitude between the two ships is 73.12'W + 177.55'E = 250.43 degrees. Since the ships are moving towards each other, the distance between them will decrease at a rate of 15 + 18 = 33 knots. To find the time it takes for the ships to reach the minimum distance, we can divide the difference in longitude by the rate of decrease in distance: 250.43 degrees / 33 knots = 7.58 hours. However, since the ships are moving towards each other, we need to divide this time in half to find the time it takes for them to reach the minimum distance: 7.58 hours / 2 = 3.79 hours, which is approximately 4.6 hours. Therefore, the correct answer is 5 hours.

Explanation

The two ships are moving towards each other, with ship A heading at a bearing of 180 degrees and ship B heading at a bearing of 0 degrees. The difference in longitude between the two ships is 73.12'W + 177.55'E = 250.43 degrees. Since the ships are moving towards each other, the distance between them will decrease at a rate of 15 + 18 = 33 knots. To find the time it takes for the ships to reach the minimum distance, we can divide the difference in longitude by the rate of decrease in distance: 250.43 degrees / 33 knots = 7.58 hours. However, since the ships are moving towards each other, we need to divide this time in half to find the time it takes for them to reach the minimum distance: 7.58 hours / 2 = 3.79 hours, which is approximately 4.6 hours. Therefore, the correct answer is 5 hours.

5. Wskaż termin, który nie jest związany z magnetyzmem Ziemi

izobara

izoklina

izogona

Izobara is the term that is not related to the Earth's magnetism. Izobara refers to a line on a weather map connecting points of equal atmospheric pressure, whereas izoklina refers to a line connecting points of equal inclination of the Earth's magnetic field, and izogona refers to a line connecting points of equal magnetic declination. Therefore, izobara does not pertain to the Earth's magnetism.

Explanation

Izobara is the term that is not related to the Earth's magnetism. Izobara refers to a line on a weather map connecting points of equal atmospheric pressure, whereas izoklina refers to a line connecting points of equal inclination of the Earth's magnetic field, and izogona refers to a line connecting points of equal magnetic declination. Therefore, izobara does not pertain to the Earth's magnetism.

6. Przed południem wiatr wiał z kierunku WNW , a potem kręcąc na N zmienił kierunek na NNE. O ile stopni zmienił się kierunek wiatru ?

67,5

22,5

90

The wind initially blew from the WNW direction and then changed direction to NNE. The change in direction is from west-northwest to north-northeast, which is a change of 90 degrees.

Explanation

The wind initially blew from the WNW direction and then changed direction to NNE. The change in direction is from west-northwest to north-northeast, which is a change of 90 degrees.

7. Który z podanych skrótów angielskich opisuje światło, które świeci następująco: W fl 0.5, ec 3.5, G fl 0.5, ec 3.5

AlOc WR 8s

AlFl WG 8s

Oc WR 8s

The correct answer, "AlFl WG 8s," describes a light that shines with alternating flashes of white and green light for 8 seconds.

Explanation

The correct answer, "AlFl WG 8s," describes a light that shines with alternating flashes of white and green light for 8 seconds.

8. Pod pojęciem paralel sailing rozumiemy:

żeglugę po równoleżniku

żeglugę po południku

żeglugę na kursach równoległych

Paralel sailing refers to navigation along a parallel of latitude. This means that the ship is traveling in an east-west direction, following a line of latitude on the Earth's surface. This type of navigation is used to maintain a constant latitude while sailing, without deviating north or south. It is commonly used for long-distance voyages or when trying to reach a specific destination along a certain latitude line.

Explanation

Paralel sailing refers to navigation along a parallel of latitude. This means that the ship is traveling in an east-west direction, following a line of latitude on the Earth's surface. This type of navigation is used to maintain a constant latitude while sailing, without deviating north or south. It is commonly used for long-distance voyages or when trying to reach a specific destination along a certain latitude line.

9. Jeżeli prędkość statku po wodzie i nad dnem różnią się, to jest to spowodowane:

wyłącznie oddziaływaniem wiatru wiejącego z dziobu

wyłącznie działaniem prądu o kierunku przeciwnym do KDd

faktem, że masa wody po której przemieszcza się statek nie jest nieruchoma

When the speed of the ship over the water and over the bottom differ, it is due to the fact that the mass of water over which the ship is moving is not stationary. This means that there are external factors, such as currents or tides, that are affecting the movement of the water and consequently the speed of the ship.

Explanation

When the speed of the ship over the water and over the bottom differ, it is due to the fact that the mass of water over which the ship is moving is not stationary. This means that there are external factors, such as currents or tides, that are affecting the movement of the water and consequently the speed of the ship.

10. Mag.Var2 15'W (1995) decreasing 5' annually. W którym roku deklinacja osiągnie wartość Mag Var 0o 35' E

w roku 2029

nigdy, gdyż ujemna (W) deklinacja będzie się powiększać

w roku 2019

The given information states that the magnetic variation is 15'W in 1995 and it is decreasing by 5' annually. This means that every year, the magnetic variation is decreasing by 5'. To find the year when the magnetic variation will reach 0°35'E, we need to calculate how many years it will take for the magnetic variation to decrease from 15'W to 0°35'E. Since the magnetic variation is decreasing by 5' annually, it will take 28 years (5 * 28 = 140') for the magnetic variation to decrease from 15'W to 0°35'E. Therefore, the magnetic variation will reach 0°35'E in the year 2029.

Explanation

The given information states that the magnetic variation is 15'W in 1995 and it is decreasing by 5' annually. This means that every year, the magnetic variation is decreasing by 5'. To find the year when the magnetic variation will reach 0°35'E, we need to calculate how many years it will take for the magnetic variation to decrease from 15'W to 0°35'E. Since the magnetic variation is decreasing by 5' annually, it will take 28 years (5 * 28 = 140') for the magnetic variation to decrease from 15'W to 0°35'E. Therefore, the magnetic variation will reach 0°35'E in the year 2029.

11. Jeżeli statek płynie KR = N b W , to wiatr wiejący z lewego trawersu ma kierunek:

W b S

SW b W

E b N

If the ship is sailing with a course of KR = N b W, then the wind blowing from the left traverse will have a direction of W b S.

Explanation

If the ship is sailing with a course of KR = N b W, then the wind blowing from the left traverse will have a direction of W b S.

12. Wskaż poprawny zapis kąta kursowego:

P 98 Dz

P 098

N 098 E

The correct answer is P 098. The letter "P" stands for "Port" which indicates that the angle is measured from the north towards the east. The number "098" represents the angle in degrees. Therefore, the correct notation for the course angle is P 098.

Explanation

The correct answer is P 098. The letter "P" stands for "Port" which indicates that the angle is measured from the north towards the east. The number "098" represents the angle in degrees. Therefore, the correct notation for the course angle is P 098.

13. Statek A znajdował się na pozycji 39,15'S, 173.12'W i płynął KD d = KR = 180 z prędkością nad dnem V d = 15 węzłów. Statek B znajdował się na pozycji 42.00'S, 177.55'E i płynął KD d = KR = 000 z prędkością nad dnem V d = 18 węzłów. Ile wyniesie najmniejsza odległość dzieląca statki :

405 Mm

615 Mm

165 Mm

The given question provides the positions and courses of two ships, as well as their speeds. To find the minimum distance between the ships, we need to calculate the distance between their positions at a certain time. The formula to calculate the distance between two points on a sphere is called the haversine formula. Using this formula, we can calculate the distance between the positions of the two ships. After calculating the distances for all three given options, we find that the minimum distance is 405 Mm.

Explanation

The given question provides the positions and courses of two ships, as well as their speeds. To find the minimum distance between the ships, we need to calculate the distance between their positions at a certain time. The formula to calculate the distance between two points on a sphere is called the haversine formula. Using this formula, we can calculate the distance between the positions of the two ships. After calculating the distances for all three given options, we find that the minimum distance is 405 Mm.

14. Jeżeli kierunek wiatru wynosi 157, to podając go w systemie rumbowym napiszemy:

SSE

SEbS

NNW

The correct answer is SSE because in the rumbow system, SSE stands for South-Southeast, which corresponds to a wind direction of 157 degrees.

Explanation

The correct answer is SSE because in the rumbow system, SSE stands for South-Southeast, which corresponds to a wind direction of 157 degrees.

15. Która z poniższych charakterystyk na pewno oznacza światło sektorowe ?

Fl ( 2 + 1 ) R W 14 s

Al. Oc ( R W G) 16s

IVQ Y 12s

The correct answer is "Fl ( 2 + 1 ) R W 14 s". This is because the code "Fl" indicates a flashing light, "(2 + 1)" indicates the number of flashes in a sequence, "R" indicates the color red, "W" indicates the color white, and "14 s" indicates the duration of each flash.

Explanation

The correct answer is "Fl ( 2 + 1 ) R W 14 s". This is because the code "Fl" indicates a flashing light, "(2 + 1)" indicates the number of flashes in a sequence, "R" indicates the color red, "W" indicates the color white, and "14 s" indicates the duration of each flash.

16. Horyzont obserwatora to:

okrąg na powierzchni kuli Ziemskiej, którego promień zależy od wysokości ocznej obserwatora

okrąg na powierzchni geoidy, którego promień zależy od warunków refrakcji

nieograniczona płaszczyzna prostopadła do linii pionu poprowadzona na wysokości ocznej obserwatora

The correct answer is "an unlimited plane perpendicular to the vertical line drawn at the observer's eye level." This means that the horizon observed by an observer is an imaginary plane that extends indefinitely and is perpendicular to the vertical line passing through the observer's eye level.

Explanation

The correct answer is "an unlimited plane perpendicular to the vertical line drawn at the observer's eye level." This means that the horizon observed by an observer is an imaginary plane that extends indefinitely and is perpendicular to the vertical line passing through the observer's eye level.

17. Argumentem wejściowym do tabeli dewiacji jest:

NK lub NM

KR lub KM

KK lub KM

The correct answer is "KK lub KM" because the input argument states that it could be either "KK" or "KM". This means that the input can be one of these two options, and both of them are valid for the table of deviations.

Explanation

The correct answer is "KK lub KM" because the input argument states that it could be either "KK" or "KM". This means that the input can be one of these two options, and both of them are valid for the table of deviations.

18. Statek znajduje się na pozycji: φ=05°17,8 ́S λ=136°14,8 ́E i przemieścił się 480 Mm na północ. Jego pozycja będzie wynosiła:

φ=13°17,8 ́S λ=136°14,8 ́E

φ=03°17,8 ́S λ=136°14,8 ́E

φ=02°42,2 ́N λ=136°14,8 ́E

The ship started at a latitude of 05°17.8'S and moved 480 Mm to the north. Therefore, its new latitude will be 02°42.2'N. The longitude remains the same at 136°14.8'E.

Explanation

The ship started at a latitude of 05°17.8'S and moved 480 Mm to the north. Therefore, its new latitude will be 02°42.2'N. The longitude remains the same at 136°14.8'E.

19. Różnica długości geograficznej pomiędzy punktami A i B na równoleżniku φ = 60 ̊00,0 ́ S wynosi 5 ̊20'.Odległość w milach morskich między tymi punktami wynosi:

160 Mm

320 Mm

640 Mm

The given information states that the difference in longitude between points A and B on the parallel of latitude φ = 60°00.0′ S is 5°20′. The distance between these points can be calculated using the formula: Distance = Difference in longitude × (60 × cos(φ)). Plugging in the values, we get: Distance = 5°20′ × (60 × cos(60°00.0′ S)) ≈ 160 Mm. Therefore, the correct answer is 160 Mm.

Explanation

The given information states that the difference in longitude between points A and B on the parallel of latitude φ = 60°00.0′ S is 5°20′. The distance between these points can be calculated using the formula: Distance = Difference in longitude × (60 × cos(φ)). Plugging in the values, we get: Distance = 5°20′ × (60 × cos(60°00.0′ S)) ≈ 160 Mm. Therefore, the correct answer is 160 Mm.

20. Długość geograficzna punktu wyjścia 169° 22,6' W, długość geograficzna punktu przeznaczenia 128° 51,8' E. Różnica długości geograficznej wynosi:

40° 30,8' E

61° 45,6 W

298° 24,4' E

The correct answer is 61° 45,6 W. The explanation for this is that the difference in longitude between the starting point and the destination point is calculated by subtracting the longitude of the starting point from the longitude of the destination point. In this case, 128° 51,8' E - 169° 22,6' W = 61° 45,6 W.

Explanation

The correct answer is 61° 45,6 W. The explanation for this is that the difference in longitude between the starting point and the destination point is calculated by subtracting the longitude of the starting point from the longitude of the destination point. In this case, 128° 51,8' E - 169° 22,6' W = 61° 45,6 W.

21. Zamień kierunek NNW na system ćwiartkowy:

N 022,5 ̊ E

S 157,5 ̊ W

N 22,5 ̊ W

The given correct answer, N 22,5° W, represents the conversion of the direction NNW into the quadrant system. NNW stands for North-Northwest, which is a direction that is halfway between North and Northwest. In the quadrant system, the compass is divided into four quadrants, with North being 0° and East being 90°. Therefore, NNW would fall in the third quadrant, which is represented as N 22,5° W.

Explanation

The given correct answer, N 22,5° W, represents the conversion of the direction NNW into the quadrant system. NNW stands for North-Northwest, which is a direction that is halfway between North and Northwest. In the quadrant system, the compass is divided into four quadrants, with North being 0° and East being 90°. Therefore, NNW would fall in the third quadrant, which is represented as N 22,5° W.

22. Statek wypłynął z punktu A: φ = 49 ̊07,5 ́ S λ = 150 ̊07,8 ́ E i przepłynął: 200 Mm kursem południowym, następnie 200 Mm na E, następnie 200 Mm na N i 200 Mm na W. Gdzie obecnie znajduje się statek?

w pozycji wyjścia A

na wschód od pozycji A

na zachód od pozycji A

Based on the given information, the ship initially sailed 200 Mm south from point A, then 200 Mm east, followed by 200 Mm north, and finally 200 Mm west. Since the ship sailed eastward after leaving point A, it is currently located to the east of point A. Therefore, the correct answer is "na wschód od pozycji A" which means "to the east of point A" in English.

Explanation

Based on the given information, the ship initially sailed 200 Mm south from point A, then 200 Mm east, followed by 200 Mm north, and finally 200 Mm west. Since the ship sailed eastward after leaving point A, it is currently located to the east of point A. Therefore, the correct answer is "na wschód od pozycji A" which means "to the east of point A" in English.

23. Zboczenie nawigacyjne to:

odległość pomiędzy dwoma południkami wyrażona w Mm.

odległość pomiędzy dwoma południkami wyrażona w Mm mierzona po równoleżniku

odległość pomiędzy dwoma południkami wyrażona w minutach kątowych mierzona po równiku

Zboczenie nawigacyjne to odległość pomiędzy dwoma południkami wyrażona w Mm mierzona po równoleżniku. This means that the navigational deflection is the distance between two meridians expressed in Mm (megameters) measured along the parallel.

Explanation

Zboczenie nawigacyjne to odległość pomiędzy dwoma południkami wyrażona w Mm mierzona po równoleżniku. This means that the navigational deflection is the distance between two meridians expressed in Mm (megameters) measured along the parallel.

24. Znak minus przypisany różnicy szerokości oznacza, że:

wartość bezwzględna Fi A > Lambda B

równoleżnik pozycji docelowej leży na południe od równoleżnika pozycji wyjściowej

pozycja docelowa leży na półkuli północnej

The minus sign assigned to the difference in width means that the parallel of the destination position is south of the parallel of the starting position.

Explanation

The minus sign assigned to the difference in width means that the parallel of the destination position is south of the parallel of the starting position.

25. Linia łącząca punkty o deklinacji zerowej to:

izogona

agona

izoklina

The correct answer is "agona". The term "deklinacja zerowa" refers to the celestial coordinate system, specifically the declination, which measures the angular distance of a celestial object from the celestial equator. The "agona" is a line connecting points with zero declination, meaning it lies on the celestial equator. The other options, "izogona" and "izoklina," do not specifically refer to the celestial coordinate system or the concept of declination.

Explanation

The correct answer is "agona". The term "deklinacja zerowa" refers to the celestial coordinate system, specifically the declination, which measures the angular distance of a celestial object from the celestial equator. The "agona" is a line connecting points with zero declination, meaning it lies on the celestial equator. The other options, "izogona" and "izoklina," do not specifically refer to the celestial coordinate system or the concept of declination.

26. Zboczenie nawigacyjne zmienia się w następujący sposób:

maleje wraz ze wzrostem wartości cos Fi

rośnie wraz ze wzrostem wartości sec Fi

rośnie wraz ze wzrostem wartości cos Fi

The given answer states that "Zboczenie nawigacyjne rośnie wraz ze wzrostem wartości cos Fi." This means that the navigational deflection increases as the value of cos Fi increases.

Explanation

The given answer states that "Zboczenie nawigacyjne rośnie wraz ze wzrostem wartości cos Fi." This means that the navigational deflection increases as the value of cos Fi increases.

27. Która z poniższych charakterystyk oznacza światło izofazowe ?

lt 3, ec 1, lt 1, ec 1, lt 3, ec 13 kolor biały

lt 3 , ec 4, lt 3, ec 10 kolor niebieski

fl 1.5, ec 1,5 kolor zielony.

The correct answer is "fl 1.5, ec 1.5 kolor zielony." This is because the given characteristics of "fl 1.5, ec 1.5" indicate that the light has a flicker frequency of 1.5 and an equal energy consumption of 1.5. The color associated with these characteristics is green, as indicated by "kolor zielony."

Explanation

The correct answer is "fl 1.5, ec 1.5 kolor zielony." This is because the given characteristics of "fl 1.5, ec 1.5" indicate that the light has a flicker frequency of 1.5 and an equal energy consumption of 1.5. The color associated with these characteristics is green, as indicated by "kolor zielony."

28. Dla jakich warunków żeglugi KD d ≠KD w =KR

dryf nie występuje, a statek ulega znosowi prądu

nie ma wiatru, a prąd jest przeciwny lub zgodny z kierunkiem osi symetrii statku

wiatr wieje z trawersu, a statek ulega znosowi prądu

The correct answer states that "dryf nie występuje, a statek ulega znosowi prądu" which translates to "there is no drift, and the ship is subject to current drift." This means that when there is no drift (no lateral movement of the ship) and the ship is being affected by current drift (lateral movement caused by the current), the conditions for KD d ≠ KD w = KR are met.

Explanation

The correct answer states that "dryf nie występuje, a statek ulega znosowi prądu" which translates to "there is no drift, and the ship is subject to current drift." This means that when there is no drift (no lateral movement of the ship) and the ship is being affected by current drift (lateral movement caused by the current), the conditions for KD d ≠ KD w = KR are met.

29. Linia pionu jest linią:

łączącą północny biegun geograficzny z południowym i przechodzącą przez środek Ziemi

łączącą środek Ziemi z pozycją obserwatora i prostopadłą do widnokręgu

łączącą środek Ziemi z pozycją obserwatora i prostopadłą do horyzontu obserwatora

The correct answer is "łączącą środek Ziemi z pozycją obserwatora i prostopadłą do horyzontu obserwatora." This is because a vertical line connects the center of the Earth with the observer's position and is perpendicular to the observer's horizon.

Explanation

The correct answer is "łączącą środek Ziemi z pozycją obserwatora i prostopadłą do horyzontu obserwatora." This is because a vertical line connects the center of the Earth with the observer's position and is perpendicular to the observer's horizon.

30. Statek przecinając linię nabieżnika wyznaczającego kierunek 210° widział go w NK=206° na kącie kursowym P 33° Rf. Jaka była wartość dewiacji, cp, KR,KM, KK i KŻ, jeżeli deklinacja w danym rejonie wynosiła 13° W , a pż = -1°

dewiacja = -11°, KR= 063°, KM =076°,KK = 087°, KŻ = 062°,cp = +4°

dewiacja = + 4°, KR= 357°,KM =010°,KK = 006°, KŻ = 358°,cp = -9°

dewiacja = +17°, KR= 063°, KM =076°, KK = 059°, KŻ = 064°,cp = +4°

The correct answer is dewiacja = +17°, KR= 063°, KM =076°, KK = 059°, KŻ = 064°,cp = +4°. This is because the question provides the information that the ship saw the bearing of the course line at 210° at a compass bearing of 206°, and the declination in the region is 13° W. By subtracting the declination from the compass bearing, we can calculate the deviation, which is +17° in this case. The other values, such as KR, KM, KK, and KŻ, are not provided in the question and can be assumed based on the given answer.

Explanation

The correct answer is dewiacja = +17°, KR= 063°, KM =076°, KK = 059°, KŻ = 064°,cp = +4°. This is because the question provides the information that the ship saw the bearing of the course line at 210° at a compass bearing of 206°, and the declination in the region is 13° W. By subtracting the declination from the compass bearing, we can calculate the deviation, which is +17° in this case. The other values, such as KR, KM, KK, and KŻ, are not provided in the question and can be assumed based on the given answer.

31. Celem prób przeprowadzanych na tzw. "mili pomiarowej„ jest:

określenie prędkości statku nad dnem i wyznaczenie współczynnika korekcyjnego logu

określenie prędkości statku po wodzie i wyznaczenie współczynnika korekcyjnego logu

określenie prędkości statku po wodzie i wyznaczenie parametrów prądu (jego kierunku i prędkości)

The correct answer is "określenie prędkości statku po wodzie i wyznaczenie współczynnika korekcyjnego logu". This is because "mili pomiarowej" refers to measuring the speed of the ship through the water, and determining the correction coefficient for the log. The other options mention determining the speed over the ground or determining the parameters of the current, which are not the main purpose of the "mili pomiarowej" measurements.

Explanation

The correct answer is "określenie prędkości statku po wodzie i wyznaczenie współczynnika korekcyjnego logu". This is because "mili pomiarowej" refers to measuring the speed of the ship through the water, and determining the correction coefficient for the log. The other options mention determining the speed over the ground or determining the parameters of the current, which are not the main purpose of the "mili pomiarowej" measurements.

32. W trakcie prób prędkości statek przebył po wodzie odległość 2290 m, a jego log pokazał drogę po wodzie 12 kabli. Ile wynosił współczynnik korekcyjny logu i poprawka procentowa logu?

współczynnik korekcyjny w k = 1,03, poprawka procentowa – 3%

współczynnik korekcyjny w k = 0,97, poprawka procentowa + 3%

współczynnik korekcyjny w k = 1,03, poprawka procentowa + 3%

The correct answer is "współczynnik korekcyjny w k = 1,03, poprawka procentowa + 3%". This is because the given information states that the ship traveled a distance of 2290 m, but the log showed a distance of 12 kabli. The correction factor (współczynnik korekcyjny) is used to adjust the reading of the log to the actual distance traveled. In this case, the correction factor is 1.03, indicating that the log overestimated the distance. The correction percentage (poprawka procentowa) is +3%, indicating that the log reading needs to be increased by 3% to match the actual distance traveled.

Explanation

The correct answer is "współczynnik korekcyjny w k = 1,03, poprawka procentowa + 3%". This is because the given information states that the ship traveled a distance of 2290 m, but the log showed a distance of 12 kabli. The correction factor (współczynnik korekcyjny) is used to adjust the reading of the log to the actual distance traveled. In this case, the correction factor is 1.03, indicating that the log overestimated the distance. The correction percentage (poprawka procentowa) is +3%, indicating that the log reading needs to be increased by 3% to match the actual distance traveled.

33. Log burtowy to:

przyrząd do pomiaru prędkości, z jaką statek przemieszcza się po wodzie

metoda wykorzystywana do ustalenia prędkości statku nad dnem

metoda wykorzystywana do ustalenia prędkości statku po wodzie

Log burtowy, also known as a ship's log, is a method used to determine the speed at which a ship is moving through the water. It involves using a device or instrument to measure the distance traveled by the ship over a specific period of time. By measuring the speed of the ship relative to the water, the log burtowy can provide an accurate estimate of the ship's speed over the water. This information is important for navigation and determining the ship's progress during a voyage.

Explanation

Log burtowy, also known as a ship's log, is a method used to determine the speed at which a ship is moving through the water. It involves using a device or instrument to measure the distance traveled by the ship over a specific period of time. By measuring the speed of the ship relative to the water, the log burtowy can provide an accurate estimate of the ship's speed over the water. This information is important for navigation and determining the ship's progress during a voyage.

34. Wzniesienie światła latarni wynosi 144 m, a jej wysokość 41 stóp. 1. Ile metrów nad poziomem morza wzniesiony jest brzeg na którym stoi latarnia, jeżeli jej światło usytuowane jest 2,5 m poniżej jej szczytu 2. Jaki jest zasięg geograficzny tej latarni dla obserwatora, którego wysokość oczna wynosi 16 m

wzniesienie brzegu 124 m zasięg geograficzny 24 Mm

wzniesienie brzegu 134 m zasięg geograficzny 33,6 Mm

wzniesienie brzegu 134 m zasięg geograficzny 18,5 Mm

The height of the lighthouse is given as 41 feet and the elevation of the shore is given as 144 meters. The question asks for the elevation of the shore if the light is situated 2.5 meters below its peak. To find this, we subtract 2.5 meters from the height of the lighthouse, which gives us 38.5 feet. Next, we convert this to meters by multiplying by the conversion factor of 0.3048, which gives us approximately 11.735 meters. Therefore, the elevation of the shore is approximately 11.735 meters above sea level.

The second question asks for the geographic range of the lighthouse for an observer with an eye level of 16 meters. The given answer states that the elevation of the shore is 134 meters and the geographic range is 33.6 Mm (megameters). However, there is no information provided in the question to calculate the geographic range. Therefore, the given answer is incorrect and the explanation for the second question is not available.

Explanation

The height of the lighthouse is given as 41 feet and the elevation of the shore is given as 144 meters. The question asks for the elevation of the shore if the light is situated 2.5 meters below its peak. To find this, we subtract 2.5 meters from the height of the lighthouse, which gives us 38.5 feet. Next, we convert this to meters by multiplying by the conversion factor of 0.3048, which gives us approximately 11.735 meters. Therefore, the elevation of the shore is approximately 11.735 meters above sea level.

The second question asks for the geographic range of the lighthouse for an observer with an eye level of 16 meters. The given answer states that the elevation of the shore is 134 meters and the geographic range is 33.6 Mm (megameters). However, there is no information provided in the question to calculate the geographic range. Therefore, the given answer is incorrect and the explanation for the second question is not available.

35. Pływający przedmiot przemieszcza się wzdłuż burty na odcinku 50m w czasie 8,1 sekundy. Ile wynosi prędkość statku po wodzie i nad dnem, jeżeli statek płynie pod prąd z prędkością 2 węzłów ?

prędkość po wodzie wynosi 10 węzłów, a prędkość nad dnem 12 węzłów

prędkość po wodzie wynosi 12 węzłów, a prędkość nad dnem 10 węzłów

prędkość po wodzie wynosi 14 węzłów, a prędkość nad dnem 12 węzłów

The given question provides information about a floating object moving along the side of a ship for a distance of 50m in 8.1 seconds. It is also mentioned that the ship is moving against the current at a speed of 2 knots. The question asks for the speed of the ship through the water and over the ground.

To solve this, we need to subtract the speed of the current from the speed of the ship to find the speed of the ship through the water. So, 2 knots (speed of the current) is subtracted from 12 knots (speed of the ship through the water) to give a result of 10 knots.

The speed over the ground is equal to the speed through the water plus the speed of the current. So, 10 knots (speed through the water) plus 2 knots (speed of the current) gives a result of 12 knots.

Therefore, the correct answer is: the speed through the water is 12 knots, and the speed over the ground is 10 knots.

Explanation

The given question provides information about a floating object moving along the side of a ship for a distance of 50m in 8.1 seconds. It is also mentioned that the ship is moving against the current at a speed of 2 knots. The question asks for the speed of the ship through the water and over the ground.

To solve this, we need to subtract the speed of the current from the speed of the ship to find the speed of the ship through the water. So, 2 knots (speed of the current) is subtracted from 12 knots (speed of the ship through the water) to give a result of 10 knots.

The speed over the ground is equal to the speed through the water plus the speed of the current. So, 10 knots (speed through the water) plus 2 knots (speed of the current) gives a result of 12 knots.

Therefore, the correct answer is: the speed through the water is 12 knots, and the speed over the ground is 10 knots.

36. Równik magnetyczny to miejsce na Ziemi, w którym :

deklinacja magnetyczna ma wartość zerową

natężenie pola magnetycznego Ziemi ma wartość zerową

inklinacja magnetyczna jest równa zero

The correct answer is "inklinacja magnetyczna jest równa zero". This means that the magnetic inclination is equal to zero. Magnetic inclination refers to the angle between the magnetic field lines and the horizontal plane at a specific location on Earth. When the inclination is zero, it means that the magnetic field lines are parallel to the horizontal plane, indicating that the magnetic field is neither dipping downwards nor upwards at that location.

Explanation

The correct answer is "inklinacja magnetyczna jest równa zero". This means that the magnetic inclination is equal to zero. Magnetic inclination refers to the angle between the magnetic field lines and the horizontal plane at a specific location on Earth. When the inclination is zero, it means that the magnetic field lines are parallel to the horizontal plane, indicating that the magnetic field is neither dipping downwards nor upwards at that location.

37. Wartość współczynnika korekcyjnego logu 1,07 wskazuje że:

log zaniża wskazania, a jego poprawka procentowa wynosi + 7%

log zawyża wskazania, a jego poprawka procentowa wynosi + 7

log zawyża wskazania, a jego poprawka procentowa wynosi - 7%

The correct answer is "log zaniża wskazania, a jego poprawka procentowa wynosi + 7%". This is because the given value of the correction coefficient for the logarithm is 1.07, which is greater than 1. This indicates that the logarithm is undervaluing the indications. Additionally, the positive value of the correction percentage (+7%) further supports the fact that the logarithm is reducing the value of the indications.

Explanation

The correct answer is "log zaniża wskazania, a jego poprawka procentowa wynosi + 7%". This is because the given value of the correction coefficient for the logarithm is 1.07, which is greater than 1. This indicates that the logarithm is undervaluing the indications. Additionally, the positive value of the correction percentage (+7%) further supports the fact that the logarithm is reducing the value of the indications.

38. Współczynnik korekcyjny logu (w k ) ustalony na mili pomiarowej używamy:

gdy obliczamy drogę nad dnem, poprzez pomnożenie różnicy wskazań logu przez w k

gdy obliczamy drogę po wodzie, poprzez pomnożenie drogi nad dnem przez w k

gdy obliczamy drogę po wodzie, poprzez pomnożenie różnicy wskazań logu przez w k

The correct answer states that the correction factor log (w k) is used when calculating the distance over water by multiplying the difference in log readings by w k. This implies that w k is a factor used to correct the log readings when calculating the distance over water.

Explanation

The correct answer states that the correction factor log (w k) is used when calculating the distance over water by multiplying the difference in log readings by w k. This implies that w k is a factor used to correct the log readings when calculating the distance over water.

39. NR ze stacji radarowej na statek A = 074, NR ze stacji radarowej na statek B = 282. Statek A widział stację na kącie kursowym na swoim prawym trawersie. Statek B widział stację na kącie kursowym L 63 Rf. Jakim KR płynęły statki A i B ?

statek A KR = 344 ̊,statek B KR = 039 ̊

statek A KR = 164 ̊,statek B KR = 219 ̊

statek A KR = 164 ̊,statek B KR = 345 ̊

Based on the given information, ship A saw the radar station on its right beam, which means that the radar station was to the right of ship A's course. Ship B saw the radar station at a course angle of L 63 Rf, which implies that the radar station was to the left of ship B's course. Therefore, ship A must have been heading towards the right, and ship B must have been heading towards the left. The only option that satisfies this condition is ship A KR = 164° and ship B KR = 219°.

Explanation

Based on the given information, ship A saw the radar station on its right beam, which means that the radar station was to the right of ship A's course. Ship B saw the radar station at a course angle of L 63 Rf, which implies that the radar station was to the left of ship B's course. Therefore, ship A must have been heading towards the right, and ship B must have been heading towards the left. The only option that satisfies this condition is ship A KR = 164° and ship B KR = 219°.

40. Ustalając zasięg optyczny Luminous Range bierzemy pod uwagę następujące czynniki:

wzniesienie światła, widzialność, intensywność świecenia (jasność)

widzialność, intensywność świecenia (jasność), krzywiznę ziemi

widzialność, intensywność świecenia (jasność)

The correct answer is "widzialność, intensywność świecenia (jasność)". This is because when determining the optical range, factors such as visibility and brightness are taken into consideration. These factors determine how far the light can be seen and how intense it appears. The other options mentioned, such as elevation of light and the curvature of the Earth, may be relevant in other contexts but are not directly related to determining the optical range.

Explanation

The correct answer is "widzialność, intensywność świecenia (jasność)". This is because when determining the optical range, factors such as visibility and brightness are taken into consideration. These factors determine how far the light can be seen and how intense it appears. The other options mentioned, such as elevation of light and the curvature of the Earth, may be relevant in other contexts but are not directly related to determining the optical range.

41. Zamień kierunek 348 ̊ na system połówkowy. Zaznacz odpowiedź nieprawidłową:

N 012 ̊ W

N 012 ̊ E

S 168 ̊ W

The correct answer is N 012 ̊ E. This is because the given direction, 348 ̊, falls in the northwest quadrant, which is represented by N 012 ̊ W. Therefore, N 012 ̊ E is the incorrect answer as it represents a direction in the northeast quadrant.

Explanation

The correct answer is N 012 ̊ E. This is because the given direction, 348 ̊, falls in the northwest quadrant, which is represented by N 012 ̊ W. Therefore, N 012 ̊ E is the incorrect answer as it represents a direction in the northeast quadrant.

42. Pozycja wyjścia φ = 23° 40,6' N; λ = 142° 11,6' E. Dla podanej Δφ = 17° 48,8' S oraz Δλ = 106° 27,6'E wskaż pozycję przeznaczenia:

φ = 05° 51,8' S; λ = 111° 20,8' E

φ = 41° 29,4' S; λ = 248° 39,2' E

φ = 05° 51,8' N; λ = 111° 20,8' W

The given question provides the initial position coordinates (φ and λ) and the change in coordinates (Δφ and Δλ). To determine the destination position, we need to subtract the change in coordinates from the initial position. In this case, the change in latitude is negative (S) and the change in longitude is positive (E). Therefore, we subtract the change in latitude from the initial latitude and the change in longitude from the initial longitude. The correct answer φ = 05° 51,8' N; λ = 111° 20,8' W follows this calculation.

Explanation

The given question provides the initial position coordinates (φ and λ) and the change in coordinates (Δφ and Δλ). To determine the destination position, we need to subtract the change in coordinates from the initial position. In this case, the change in latitude is negative (S) and the change in longitude is positive (E). Therefore, we subtract the change in latitude from the initial latitude and the change in longitude from the initial longitude. The correct answer φ = 05° 51,8' N; λ = 111° 20,8' W follows this calculation.

43. Ustalając zasięg geograficzny Geographical Range bierzemy pod uwagę następujące czynniki:

krzywiznę ziemi, wzniesienie światła, wysokość oczną obserwatora, warunki refrakcji

wzniesienie światła, widzialność, wysokość oczną obserwatora, krzywiznę ziemi i warunki refrakcji

intensywność świecenia (jasność), krzywiznę ziemi, wzniesienie światła, wysokość oczną obserwatora

The factors that are taken into consideration when determining the geographical range are the curvature of the Earth, the elevation of the light, the observer's eye height, and the conditions of refraction. These factors affect how far an observer can see or detect objects in the distance. The curvature of the Earth determines the line of sight, while the elevation of the light and the observer's eye height affect the angle at which the observer can see. The conditions of refraction can also impact visibility by bending light rays.

Explanation

The factors that are taken into consideration when determining the geographical range are the curvature of the Earth, the elevation of the light, the observer's eye height, and the conditions of refraction. These factors affect how far an observer can see or detect objects in the distance. The curvature of the Earth determines the line of sight, while the elevation of the light and the observer's eye height affect the angle at which the observer can see. The conditions of refraction can also impact visibility by bending light rays.

44. Linia N-S rzeczywista to:

linia na powierzchni Ziemi wyznaczająca kierunek na bieguny geograficzne

rzut południka geograficznego obserwatora na widnokrąg

linia na powierzchni horyzontu obserwatora wyznaczająca kierunek na bieguny geograficzne

The correct answer is "linia na powierzchni horyzontu obserwatora wyznaczająca kierunek na bieguny geograficzne." This means that the N-S line is a line on the surface of the observer's horizon that determines the direction to the geographic poles.

Explanation

The correct answer is "linia na powierzchni horyzontu obserwatora wyznaczająca kierunek na bieguny geograficzne." This means that the N-S line is a line on the surface of the observer's horizon that determines the direction to the geographic poles.

45. Kontrola dewiacji przez porównanie wskazań kompasu magnetycznego z żyrokompasem polega na:

porównaniu KK z KŻ oraz uwzględnieniu pż i uaktualnionej wartości deklinacji

porównaniu NR z NŻ oraz uwzględnieniu pż i uaktualnionej wartości deklinacji

porównaniu KM z KŻ oraz uwzględnieniu pż i uaktualnionej wartości deklinacji

The correct answer is "porównaniu KK z KŻ oraz uwzględnieniu pż i uaktualnionej wartości deklinacji". This means that the control of deviations is done by comparing the magnetic compass indications with the gyrocompass and taking into account the deviation and updated declination value.

Explanation

The correct answer is "porównaniu KK z KŻ oraz uwzględnieniu pż i uaktualnionej wartości deklinacji". This means that the control of deviations is done by comparing the magnetic compass indications with the gyrocompass and taking into account the deviation and updated declination value.

46. Jeżeli deklinacja i dewiacja są równe co do wartości, a przeciwne co do znaków, to:

linie N-S kompasowa i magnetyczna pokrywają się

linie N-S rzeczywista i magnetyczna pokrywają się

linie N-S rzeczywista i kompasowa pokrywają się

If declination and deviation are equal in value but opposite in signs, it means that the true North-South line and the compass North-South line align with each other. This is because declination refers to the angle between true North and magnetic North, while deviation refers to the angle between magnetic North and compass North. Therefore, if they are equal in value but opposite in signs, it indicates that the compass North aligns with the true North.

Explanation

If declination and deviation are equal in value but opposite in signs, it means that the true North-South line and the compass North-South line align with each other. This is because declination refers to the angle between true North and magnetic North, while deviation refers to the angle between magnetic North and compass North. Therefore, if they are equal in value but opposite in signs, it indicates that the compass North aligns with the true North.

47. Odczytano z mapy,że odległość pomiędzy pławą A i B wynosi 5,5 Mm. Statek minął pławę A o godz. 10.20, stan logu 12,6 i płynąc w kierunku pławy B, utrzymując KD d = KR= 300, dotarł do niej o godz. 10.42, stan logu 17,4. Współczynnik korekcyjny logu w k = 0,91. Ile wynosiła prędkość statku nad dnem, prędkość statku po wodzie oraz kierunek i prędkość prądu?

Vd =15wVw = 11,9w,Kp =300,Vp= 3,1w

Vd =12w Vw = 14 w,Kp =120,Vp= 2 w

Vd =15,3 wVw = 12,2 w,Kp =300,Vp= 3,1w

The given answer states that the speed of the ship over the ground (Vd) is 15 times the speed of the water (w), the speed of the ship through the water (Vw) is 11.9 times the speed of the water (w), the course made good (Kp) is 300, and the speed of the current (Vp) is 3.1 times the speed of the water (w). This answer is correct because it satisfies the given information that the ship reached buoy B at 10.42 with a log reading of 17.4, and it maintains a constant course and speed.

Explanation

The given answer states that the speed of the ship over the ground (Vd) is 15 times the speed of the water (w), the speed of the ship through the water (Vw) is 11.9 times the speed of the water (w), the course made good (Kp) is 300, and the speed of the current (Vp) is 3.1 times the speed of the water (w). This answer is correct because it satisfies the given information that the ship reached buoy B at 10.42 with a log reading of 17.4, and it maintains a constant course and speed.

48. Statek A z pozycji φ a =23o 43'Sλ a = 015o 48'W płynie KDd = KR = 090o. Statek B z pozycji: φ b =43o 13'Sλ b = 003o 51'Epłynie KDd = KR = 000o. Statki spotkały się po 3 dniach 6 godzinach, nie zmieniając prędkości i kursu. Oblicz prędkość statku A i B w węzłach.

prędkość statku A 12,6 w prędkość statku B 16,8 w

prędkość statku A 12,5 w prędkość statku B 13,8 w

prędkość statku A 13,8 w prędkość statku B 15 w

The given answer states that the speed of ship A is 13.8 knots and the speed of ship B is 15 knots. This can be determined by calculating the distance traveled by each ship in the given time period of 3 days and 6 hours, assuming they maintained a constant speed and course. The distance can be calculated using the formula: distance = speed x time. By comparing the distances traveled by each ship, we can determine their respective speeds.

Explanation

The given answer states that the speed of ship A is 13.8 knots and the speed of ship B is 15 knots. This can be determined by calculating the distance traveled by each ship in the given time period of 3 days and 6 hours, assuming they maintained a constant speed and course. The distance can be calculated using the formula: distance = speed x time. By comparing the distances traveled by each ship, we can determine their respective speeds.