Physics: A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is exposed to a medium at 50°F, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/h·ft2·°F. Calculate the critical radius (rcr) of the plastic insulation (in inches).

A 0.091-in-diameter electrical wire at 90°F is covered

linhlema2005

02.02.2022

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

$r_{cr}=\dfrac{2K}{h}$

Where, k = Material conductivity

h = Heat transfer coefficient

Put the value into the formula

$r_{cr}=\dfrac{2\times0.075}{2.5}$

$r_{cr}=0.06\ ft$

$r_{cr}=0.72\ inches$

Hence, The critical radius of the plastic insulation is 0.72 inches.

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kekecantonxox121

12.10.2021

Explanation:

It is given that,

Initial velocity of the truck, u = 25 m/s

Final speed of the truck, v = 40 m/s

Time taken, t = 3 s

We need to find the acceleration of the truck. It can be calculated using formula as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{40-25}{3}$

$a=5\ m/s^2$

So, the acceleration of the truck is 5 meter per second square. Hence, this is the required solution.

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