If the speed necessary for reaching port at a... - USCG Navigation Problems

If the speed necessary for reaching port at a designated time is 23.7 knots and the pitch of the propeller is 20.8 feet, how many revolutions per minute will the shaft have to turn, assuming a 7% negative slip?

A 124

B 108

C 112

D 116

AI Explanation

The correct answer is B) 108 revolutions per minute. To calculate the required shaft revolutions per minute, we can use the formula: Shaft RPM = Speed / (Pitch * (1 - Slip)) Given information: - Speed = 23.7 knots - Pitch = 20.8 feet - Slip = 7% Plugging in the values: Shaft RPM = 23.7 / (20.8 * (1 - 0.07)) Shaft RPM = 23.7 / (20.8 * 0.93) Shaft RPM = 108 The other options are incorrect because they do not correctly apply the formula and account for the 7% negative slip.

Related Questions

Q@q.QuestionNumber:On 16 January at 1804 zone time, you take a sextant observation of Polaris. Your... Q@q.QuestionNumber:On 24 March your vessel is enroute from Cadiz to Norfolk. Evening twilight will ... Q@q.QuestionNumber:On 22 June your 0424 DR position is LAT 26°18.5' N, LONG 124°18.2' W. You observ... Q@q.QuestionNumber:On 2 October your 1845 DR position is LAT 28°09.2'S, LONG 167°48.1'E. You observ... Q@q.QuestionNumber:On 22 July your 0442 DR position is LAT 26°35.6' N, LONG 22°16.7' W. You observe...

Ready to test your knowledge?

Take a Navigation Problems Practice Exam

Official Resources

USCG Navigation Rules (COLREGS) 33 CFR Navigation Rules USCG National Maritime Center

Want to practice with timed exams?

Download Coast Guard Exams for the full exam experience.

Get the App

Question 65Navigation Problems

If the speed necessary for reaching port at a designated time is 23.7 knots and the pitch of the propeller is 20.8 feet, how many revolutions per minute will the shaft have to turn, assuming a 7% negative slip?

Want to practice with timed exams?