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3 years ago

9

Input signal to a controller is

1 answer:

alexgriva [62]3 years ago

4 0

Answer:

were the cord plugs in

Explanation:

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Suppose that a class CalendarDate has been defined for storing a calendar date with month, day and year components. (In our sect

laiz [17]

Answer:

note:

find the attachment

7 0

3 years ago

Do the coil resistances have any effect on the plots?

PolarNik [594]

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.

8 0

2 years ago

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

dybincka [34]

Answer:153.76 MPa

Explanation:

$Initial Area\left ( A_0\right )=12.56 mm^2$

$Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2$

$Die angle=30^{\circ}$

$\alpha =\frac{30}{2}=15^{\circ}$

$\mu =0.08$

$Yield stress\left ( \sigma _y \right )=350 MPa$

$B=\mu cot\left ( \aplha\right )=0.2985$

$\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B$

$\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}$

$\sigma _{pressure}=153.76 MPa$

8 0

2 years ago

1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

5 0

3 years ago

A 200-mm-long strip of metal is stretched in two steps, first to 300 mm and then to 400 mm. Show that the total true strain is t

Neko [114]

Explanation:

For true Strain:

step 1:

E true = Ln(1 + 0.5 ) = 0.40

Step 2:

E true = Ln(1 + 0.33 ) = 0.29

By single step process:

E true = Ln(1 + 1 ) = 0.69

total strain of step process = 0.40 + 0.29 = 0.69 units

SO TRUE STRAIN IS ADDITIVE.

4 0

2 years ago