True or False: Global wind patterns have less to do with air temperature than local winds.
1 answer:
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Answer:
The critical radius of the plastic insulation is 0.72 inches.
Explanation:
Given that,
Diameter = 0.091 in
Thickness = 0.02 in
Initial temperature = 90°F
Final temperature = 50°F
Heat transfer coefficient = 2.5 Btu/h.ft²°F
Material conductivity = 0.075 Btu/h.ft °F
We need to calculate the critical radius of the plastic insulation
Using formula of critical radius
Where, k = Material conductivity
h = Heat transfer coefficient
Put the value into the formula
Hence, The critical radius of the plastic insulation is 0.72 inches.
Refer to the diagram shown below.
The net force acting on the box is 17 - 13 = 4 N to the right.
The box moves on a friction surface by 3.5 m to the right.
By definition,
Work = Force x Distance.
(a) The work done by the girl is
W₁ = (17 N)*(3.5 m) = 59.5 J
(b) The work done by the boy is
W₂ = (13 N)*(-3.5 m) = - 45.5 J
(c) The work done by the net force is
W₃ = (4 N)*(3.5 m) = 14 J
Note that W₃ = W₁ + W₂
Answers:
(a) 59.5 J
(b) - 45.5 J
(c) 14 J
The net force acting on the box is 17 - 13 = 4 N to the right.
The box moves on a friction surface by 3.5 m to the right.
By definition,
Work = Force x Distance.
(a) The work done by the girl is
W₁ = (17 N)*(3.5 m) = 59.5 J
(b) The work done by the boy is
W₂ = (13 N)*(-3.5 m) = - 45.5 J
(c) The work done by the net force is
W₃ = (4 N)*(3.5 m) = 14 J
Note that W₃ = W₁ + W₂
Answers:
(a) 59.5 J
(b) - 45.5 J
(c) 14 J