A composite wall consists of 20 mm thick steel plate backed by insulation brick (k = 0.39 W/mK) of 50 cm thickness and overlaid

by mineral wool of 20 cm thickness (k = 0.05 W/mK) and 70 cm layer of brick of (k = 0.39 W/mK). The inside is exposed to convection at 650°C with h = 65 W/ m2K. The outside is exposed to air at 35°C with a convection coefficient of 15 W/m2K. Determine the heat loss per unit area, interface temperatures and temperature gradients in each materials.

Engineering

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All MOS devices are subject to damage from:________

Viktor [391]

Answer:

d. all of these

Explanation:

Electrostatic discharge typically generates a high voltage in a localized area, leading to increased current and heat. It can create craters in an MOS device, damage bond wires, or result in other forms of impairment. Essentially, MOS devices are vulnerable to harm from "all of these."

6 0

2 months ago

A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible m

choli [298]

Answer:

a) v₂ = 26.6 ft/s

b) v₂ = 31.9 ft/s

Explanation:

a) Applying the principle of energy conservation for the soccer arm from the point of impact to its lowest point:

$mgh=\frac{1}{2} mv^{2} \\v=\sqrt{2gh} =\sqrt{2*32.2*6} =19.66ft/s$

The ball's velocity in the tangential direction is given by:

v₂*sinθ = v₁*sin30

With the values:

if v₁ = 0

θ = 0º

The restitution coefficient is:

$e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{1}*cos30 )} \\0.8=\frac{v_{2}cos0-v_{2}*cos30 }{19.66*cos30-(0*cos30)} \\v_{2} =\frac{v_{2}-13.6 }{cos30}$

where O=θ

The aggregate momentum equation is:

$mv-mv_{1} =mv_{2} +mv_{2} cos(O+30)\\$ , where O = θ

$mv-mv_{1} =m(\frac{v_{2}-13.6 }{cos30} )+mv_{2} cos(O+30)$

When we incorporate gravitational acceleration:

$mgv-mgv_{1} =mg(\frac{v_{2}-13.6 }{cos30} )+mgv_{2} cos(O+30)\\5*19.66-1*0=5*(\frac{v_{2}-13.6 }{cos30} )+1*v_{2} cos(O+30)$

Isolating v₂ results in:

v₂ = 26.6 ft/s

b) To find the ball's velocity, we utilize:

v₂ * sinθ = v₁ * sin30

if v₁ = 10 ft/s

then v₂ * sinθ = 10 * sin30

thus sinθ = 5/v₂

The restitution coefficient remains:

$e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{2}*cos30) } \\0.8=\frac{v_{2}cosO-v_{2}cos30 }{19.66*cos30-(-10*cos30)} \\v_{2} =\frac{v_{2}cos30-20.55}{cos30}$