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°
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°
College Physics
1st Edition
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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Topic Video
Question
ASAP
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f621d1d-a275-4cc5-b309-3478432e3f02%2F00f48301-8477-46dd-944c-d2b117f1ad28%2Fv2zmnq_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f621d1d-a275-4cc5-b309-3478432e3f02%2F00f48301-8477-46dd-944c-d2b117f1ad28%2Fjyy5xl_processed.jpeg&w=3840&q=75)
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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