The electrical panel schedules are located on EWR Plan number __

140-character statement that completes this sentence: I pledge to not text and drive because...

s2008m [1.1K]

Answer:

its bad and to drive drunk is even worse

4 0

2 years ago

Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i

leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

$R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}$

$R_{convection}=\frac{1}{h(2\pi r_{2}L)}$

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0

2 years ago

An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30

sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

$F = LwY_{avg}$

where;

w is the width of the annealed copper

$Y_{avg}$ is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

$L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm$

Now, determine the average true stress, $Y_{avg}$ , for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

$F = LwY_{avg}$

$F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN$

The power required in this operation is given by;

$P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW$

5 0

2 years ago

Tech A says that as moisture levels increase in brake fluid, the boiling point of the brake fluid decreases. Tech B says that fi

dezoksy [38]

Answer: Both Technician A and B are correct.

Explanation:

Technicians A and B are both right about their diagnosis. The Society of Automotive Engineers performed extensive research on vehicle brake fluids and found that there is typically a 2% moisture content in the brake fluid after a year of operating a vehicle. And as the moisture content of the brake fluid rises, the boiling point of the brake fluid decreases as well.

4 0

2 years ago

A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$