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3 years ago

What the minimum wire size for a general residential application on a 20 A circuit

Engineering

1 answer:

prisoha [69]3 years ago

7 0

Answer:

The minimum wire size for a 20 A circuit would be 14 gauge wire. However modern fire and building (in the United States) code dictates that 12 gauge wire be used for 20 A. This is for added safety because there is a chance that 14 gauge wire can be overloaded by 20 A if the situation is right.

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The inverted U-tube is used to measure the pressure difference between two points A and B in an inclined pipeline through which

JulijaS [17]

Answer:

i) 0.610 m or 610 mm

ii) 0.4 m or 400 mm

Explanation:

The pressure difference between the pipes is

a) Air

Pa + πha +Ha = Pb + πhb +Hb

Pa - Pb = π(hb-ha) + Hb-Ha

Relative density of air = 1.2754 kg /m3

Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm

b) paraffin of relative density of 0.75

Pa - Pb = π(hb-ha) + Hb-Ha

Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm

8 0

3 years ago

CAN YALL GO FOLLOW MY INSTAGRAM PLSS

lions [1.4K]

Answer:

what is it

Explanation:

7 0

3 years ago

Read 2 more answers

Assume that each atom is a hard sphere with the surface of each atom in contact with the surface of its nearest neighbor. Determ

deff fn [24]

Answer:

The classification of the concern is listed in the interpretation segment below.

Explanation:

(a)...

Simple cubic lattice

$a=2r$

Now,

The unit cell volume will be:

$=a^3$

$=(2r)^3$

$=8r^3$

At one atom per cell, atom volume will be:

$=(1)\times (\frac{4 \pi r^3}{3})$

Then the ratio will be:

$Ratio=\frac{\frac{4 \pi r^3}{3}}{8r^3}\times 100 \ percent$

$=52.4 \ percent$

(b)...

Diamond lattice

The body diagonal will be:

$d=8r=a\sqrt{3}$

$a=\frac{8}{\sqrt{3}}r$

The unit cell volume will be:

$=a^1$

$=(\frac{8r}{\sqrt{3}})^1$

At eight atom per cell, the atom volume will be:

$=8(\frac{4 \pi r^1}{3})$

Then the Ratio will be:

$Ratio=\frac{8(\frac{4 \pi r^1}{3})}{(\frac{8r}{\sqrt{3}})^1}\times 100 \ percent$

$=34 \ percent$

Note: percent = %

5 0

3 years ago

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

aksik [14]

Answer:

48087.5MPa

Explanation:

The formula for this is

σ = 2σo√L/R

Using the information below we can solve this problem

Crack length L = 0.37mm

Radius R = 0.001 = 1x10^-3

Stress σo = 1250MPa

When we insert values into the formula

σ = 2(1250)√0.37/0.001

= 2(1250)√370

= 2(1250)(19.235)

= 48087.5 Mpa

This is therefore the theoretical fracture strength of the brittle material.

4 0

3 years ago

Using bearings will cause more friction. A.True B.False

anyanavicka [17]

(False) bearings reduce friction by using smooth balls lubricated with oil or grease that freely roll between a smooth inner and outer surface. The main concept of the ball bearing is that objects that roll past each other produce less friction than if the objects were sliding against each other.

5 0

3 years ago