problem06_94

# University Physics with Modern Physics with Mastering Physics (11th Edition)

This preview shows page 1. Sign up to view the full content.

6.94: a) The number of cars is the total power available divided by the power needed per car, , 177 m/s) N)(27 10 8 . 2 ( W 10 4 . 13 3 6 = × × rounding down to the nearest integer. b) To accelerate a total mass M at an acceleration a and speed v , the extra power needed is Mav . To climb a hill of angle α , the extra power needed is . sin v Mg These will be nearly the same if ; sin ~ g a if , m/s 10 . 0 ~ tan ~ sin 2 g g the power is about the same as that needed to accelerate at .
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: m/s 10 . 2 c) MW. 2.9 m/s) 27 )( 010 . )( m/s kg)(9.80 10 10 . 1 ( 2 6 = × d) The power per car needed is that used in part (a), plus that found in part (c) with M being the mass of a single car. The total number of cars is then , 36 m/s) 27 ))( 010 . )( m/s kg)(9.80 10 (8.2 N 10 8 . 2 ( W 10 9 . 2 W 10 4 . 13 2 4 3 6 6 = × + × ×-× rounding to the nearest integer....
View Full Document

## This document was uploaded on 02/04/2008.

Ask a homework question - tutors are online