Q4/ For the figure below:*a) EFy is:0 0 0 0 0 0 C0.6m0 0 0 0 0 0 C0.2O-860.12F2=1000 N-824.59-831.640 0 0 0 0 0 0-900-789.22- 840.88Q4/b) EFx is: *Other:O-525.146O-521.46-510.2490.54-499.54-515.320 Other:Q4/C) The resultant R is *977.6983.57950.21945.54960.24965.22400Other:0.8mF1=1600N+0.44F3=900 N3F4=1200NQ4/d) The resultant direction is0 0 0 0 0 0 0 058606554505263Other:

Given:The following forces are acting on a body:F1=1600 NF2=1000 NF3=900 NF4=1200 N

Answered: Q4/ For the figure below: a) Fy is:…

Engineering

Mechanical Engineering

Q4/ For the figure below: a) Fy is: 0.6m 0.2 40° 0.8m FI-1600 N 0.4

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.2P: 5.2 Evaluate the Prandtl number from the following data: , .

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Question

Q4/ For the figure below:*
a) EFy is:
0 0 0 0 0 0 C
0.6m
0 0 0 0 0 0 C
0.2
O-860.12
F2=1000 N
-824.59
-831.64
0 0 0 0 0 0 0
-900
-789.22
- 840.88
Q4/b) EFx is: *
Other:
O-525.146
O-521.46
-510.2
490.54
-499.54
-515.32
0 Other:
Q4/C) The resultant R is *
977.6
983.57
950.21
945.54
960.24
965.22
400
Other:
0.8m
F1=1600N
+0.4
4
F3=900 N
3
F4=1200N
Q4/d) The resultant direction is
0 0 0 0 0 0 0 0
58
60
65
54
50
52
63
Other:

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