the bottom row 2nd box selections are [up/down] Question 1P1 = 5.4 kg m/s. 32.9° south of upP2 = 13.9 kg m/s, 33.4° south of downPi +p, = [Select ][ Select ]kg m/s, [ Select ][ Select ][ Select ](relative to up/down)Question 2F1 = 8.6 N. 29.7° below southF, = 13.7 N, 22.8° south of upF, + F2 =[ Select ][ Select ]N.[ Select ][ Select ](relative to up/down)Question 3di = 7.9 m. 38.4° west of downd2 = 11.9 m. 14.5° west of down[ Select ][ Select]m,-[ Select ][ Select ](relative to up/down)> [ Select ][ Select ](relative to up/down)[ Select ]east ofwest ofnorth ofsouth ofabovebelow

Answered: Question 1 P1 = 5.4 kg m/s. 32.9° south…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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the bottom row 2nd box selections are [up/down]

Question 1
P1 = 5.4 kg m/s. 32.9° south of up
P2 = 13.9 kg m/s, 33.4° south of down
Pi +p, = [Select ]
[ Select ]
kg m/s, [ Select ]
[ Select ]
[ Select ]
(relative to up/down)
Question 2
F1 = 8.6 N. 29.7° below south
F, = 13.7 N, 22.8° south of up
F, + F2 =
[ Select ]
[ Select ]
N.
[ Select ]
[ Select ]
(relative to up/down)
Question 3
di = 7.9 m. 38.4° west of down
d2 = 11.9 m. 14.5° west of down
[ Select ]
[ Select]
m,
-
[ Select ]
[ Select ]
(relative to up/down)
>

[ Select ]
[ Select ]
(relative to up/down)
[ Select ]
east of
west of
north of
south of
above
below

Expert Solution

Step 1

As per the guide lines first question is solved, kindly post other question as new post.

Question 1

Consider south as $\hat{i}$ direction, up as $\hat{j}$ and down as $- \hat{j}$ direction.

South of up means the vector is in south - up plane, the angle is measured from up towards south.

South of down means the vector is in south - down plane, the angle is measured from down towards south.

${\vec{p}}_{1} = (5.4 kg \frac{m}{s}) 32.9 ° south of up = (5.4 kg \frac{m}{s}) [\sin 32.9 ° \hat{i} + \cos 32.9 ° \hat{j}] = (2.93314203 \hat{i} + 4.533947268 \hat{j}) kg \frac{m}{s} {\vec{p}}_{2} = (13.9 kg \frac{m}{s}) 33.4 ° south of down = (13.9 kg \frac{m}{s}) [\sin 33.4 ° \hat{i} - \cos 33.4 ° \hat{j}] = (7.651682287 \hat{i} - 11.6043853 \hat{j}) kg \frac{m}{s}$

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