{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PHY 303K - Homework 12

# PHY 303K - Homework 12 - homework 12 – FIERRO JEFFREY –...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 12 – FIERRO, JEFFREY – Due: Mar 1 2008, 11:00 pm 1 Question 1, chap 8, sect 5. part 1 of 1 10 points An outboard motor propels a boat through the water at 13 mi / h. The water resists the forward motion of the boat with a force of 23 lb. How much power is produced by the motor? Correct answer: 0 . 797334 hp (tolerance ± 1 %). Explanation: First, let us change the velocity to ft/s. = (13 mi / h) · parenleftbigg 5280 ft 1 s parenrightbigg · parenleftbigg 1 hr 3600 s parenrightbigg = 19 . 0667 ft / s . Since the boat moves at a constant velocity, the power output of its motor is P = F v u = (23 lb) (19 . 0667 ft / s) = (438 . 534 ft lb / s) parenleftbigg 1 hp 550 ft lb / s parenrightbigg = 0 . 797334 hp . Question 2, chap 8, sect 5. part 1 of 2 10 points A 1990 kg car starts from rest and acceler- ates uniformly to 18 m / s in 16 . 2 s . Assume that air resistance remains con- stant at 307 N during this time. Find the average power developed by the engine. Correct answer: 30 . 3794 hp (tolerance ± 1 %). Explanation: m = 1990 kg , v i = 0 m / s , v f = 18 m / s , and Δ t = 16 . 2 s . The acceleration of the car is a = v f- v i Δ t = v f Δ t since v i = 0, so a = 18 m / s 16 . 2 s = 1 . 11111 m / s 2 . Thus the constant forward force due to the engine is found from summationdisplay F = F engine- F air = ma F engine = F air + ma = 307 N + (1990 kg) ( 1 . 11111 m / s 2 ) = 2518 . 11 N . The average velocity of the car during this interval is v av = v f + v i 2 , so the average power output is P = F engine v av = F engine parenleftBig v f 2 parenrightBig = (2518 . 11 N) parenleftbigg 18 m / s 2 parenrightbiggparenleftbigg 1 hp 764 W parenrightbigg = 30 . 3794 hp . Question 3, chap 8, sect 5. part 2 of 2 10 points Find the instantaneous power output of the engine at t = 16 . 2 s just before the car stops accelerating. Correct answer: 60 . 7587 hp (tolerance ± 1 %). Explanation: The instantaneous velocity is 18 m / s and the instantaneous power output of the engine is P = F engine v f = (307 N)(18 m / s) parenleftbigg 1 hp 764 W parenrightbigg = 60 . 7587 hp . homework 12 – FIERRO, JEFFREY – Due: Mar 1 2008, 11:00 pm 2 Question 4, chap 8, sect 2....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

PHY 303K - Homework 12 - homework 12 – FIERRO JEFFREY –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online