T-30 = Cekt T = Cekt + 30 T(10) = 0°F: 0 Ce¹0k + 30 (1) = T(20) = 15°F: = 15 Ce 20k + 30 (2) k = -0.06931 C = -60.00 T = 30 - 60.00e-0.06931 t B] T(0) =? T = 30-60e-0.06931 (0) T(0) = -30.00°F
T-30 = Cekt T = Cekt + 30 T(10) = 0°F: 0 Ce¹0k + 30 (1) = T(20) = 15°F: = 15 Ce 20k + 30 (2) k = -0.06931 C = -60.00 T = 30 - 60.00e-0.06931 t B] T(0) =? T = 30-60e-0.06931 (0) T(0) = -30.00°F
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
explain how did it get the value of c and k
![T-30 = Cekt
T = Cekt + 30
T(10) = 0°F:
0 Ce¹0k + 30 (1)
=
T(20) = 15°F:
15 Ce 20k + 30 (2)
=
k = -0.06931
C = -60.00
T = 30 - 60.00e-0.06931 t
B] T(0) =?
T = 30-60e-0.06931 (0)
T(0) = -30.00°F](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07b872e5-00cb-432d-bbe2-49f1110b4c2e%2F5002e18a-07e1-4f89-9bdb-05159a83f2eb%2Fnkp36yh_processed.png&w=3840&q=75)
Transcribed Image Text:T-30 = Cekt
T = Cekt + 30
T(10) = 0°F:
0 Ce¹0k + 30 (1)
=
T(20) = 15°F:
15 Ce 20k + 30 (2)
=
k = -0.06931
C = -60.00
T = 30 - 60.00e-0.06931 t
B] T(0) =?
T = 30-60e-0.06931 (0)
T(0) = -30.00°F
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