Dala 78 = 100 g, = = 75 91 C = 12591 18° S of W 5 = 50g, 80.5 S of Et be of Req'd pragy [A + B += €] B = - E Sol'n: Vector 14 100 10 14 7.5° N of E 80° N of E B = tan X 100 cos (7.5°) 75 cos (80°) cos (18°) 6.713965074 - 125 2 (-6.713965074)2 + (48.286 0764)² TÈL = TEL= 48.75061539 9 48.2860764 -6.713965074 O Y 100 sin (7.5°) 75 sin (90°) - 125 sin (18⁰) 48. 2860764 = -82, 08403014. Final Answer: √ F1 = 50g € = -20°

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ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter13: Temperature, Kinetic Theory, And The Gas Laws
Section: Chapter Questions
Problem 11PE: What is the change in length of a 3.00mlong column of mercury if its temperature changes from 37.0C...
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JE!
Data
À=
= 100 g,
7.5° N
80° N of E
S of W
550g, 80.5 5 of Et
B =
759
C =
= 125g
Sol'n:
Req'd prago
[A + B + C = €] @ ₂
6 = - Eª
DE-
TEL=
18°
1
Vector
2
B
Ĉ
of E
S bec
tan`
X
100 cos (7.5°)
75 cos (80°)
- 125 cos (18⁰)
-6.713965074
48.75061539
(-6.713965074)2 + (48.2860764)²
g
48.2860764
- 6.713965074.
= -82, 08403014
DATE.
O
Y
100 sin (7.5°)
75 sin (90°)
- 125 sin (18⁰)
48. 2860764
Final Answer:
1 F1 = 50g
€ = = -20°
80
Transcribed Image Text:JE! Data À= = 100 g, 7.5° N 80° N of E S of W 550g, 80.5 5 of Et B = 759 C = = 125g Sol'n: Req'd prago [A + B + C = €] @ ₂ 6 = - Eª DE- TEL= 18° 1 Vector 2 B Ĉ of E S bec tan` X 100 cos (7.5°) 75 cos (80°) - 125 cos (18⁰) -6.713965074 48.75061539 (-6.713965074)2 + (48.2860764)² g 48.2860764 - 6.713965074. = -82, 08403014 DATE. O Y 100 sin (7.5°) 75 sin (90°) - 125 sin (18⁰) 48. 2860764 Final Answer: 1 F1 = 50g € = = -20° 80
DATE.
1
% Difference of IDI and TF1 =
% Difference of
(94)
K
and E
50-50
-
(23.8) 200 00
(²68) 200 25
50
3
80.5-(-80)
-80.5
0.62111801248
0:6%
PROZOPO 15-07
Tal
Transcribed Image Text:DATE. 1 % Difference of IDI and TF1 = % Difference of (94) K and E 50-50 - (23.8) 200 00 (²68) 200 25 50 3 80.5-(-80) -80.5 0.62111801248 0:6% PROZOPO 15-07 Tal
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