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3 years ago

The A+B = A + B then angle

Physics

1 answer:

saw5 [17]3 years ago

6 0

Answer:

sección B. Sistemas de numeración (16 problemas). 21. Sección C ... 3. A 12- Hallar el valor exacto de la expresión. 4,.. 5 2,.. 7 0,45.. 3 ... 60. 390,625kg. El importe que supone dicha economía en las 90. 14 ... pueden formarse con las cinco vocales y diez consonantes, cinco de éstas serán b, c, d, f, g

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A cylindrical tungsten filament 16.0 cm long with a diameter of 1.00 mm is to be used in a machine for which the temperature wil

KonstantinChe [14]

Answer:

Explanation:

Resistance of the tungsten wire

R = resistivity x length / cross sectional area

= $\frac{5.25\times10^{-8}\times16\times10^{-2}}{3.14\times(.5\times10^{-3})^2}$

= 107 x 10⁻⁴ ohm

Resistance at 120 degree can be obtained from the following formula

$R_t = R_0( 1 + \alpha\times t )$

$R_t = 107\times10^{-4}( 1 + .0045\times 100)$

= 155.15 x 10⁻⁴ ohm

= 160 x 10⁻⁴ ohm ( rounding off to two syg fig )

current = 12.5

potential diff = 12.5 x 155.15 x 10⁻⁴ V

= 0 .1939 V

= .19 V

required electric field = potential diff / length of wire

= .1939 / 16 x 10⁻²

= 1.2 N / C

4 0

3 years ago

Read 2 more answers

Kindly Don't Spàm!<br>Thank uh !:)<br><img src="https://tex.z-dn.net/?f=%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20" id="Te

olga55 [171]

$\: \underline{ \boxed{ { \color{gray}\frak{\huge \: Answer : }}}}$

<h3> Collector Current at Saturation :</h3>

$\sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }$

$\: \:$

$\sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }$

$\: \:$

$\bold{ \bf \large {I_{C(SAT)} =5.45mA }}$

$\: \:$

________________________________

<h3> Value Of Cut - off Voltage : </h3>

$\sf \large \: V_{CS(cut-off)} = V_{CC}$

$\: \:$

Therefore ,

$\: \:$

$\bf \large \: V_{CS(cut-off)} = \: 12v$

$\: \:$

________________________________

${ \bf\large \: Base \: Current , I_B = \frac{V_{CC}}{R_C} }$

$\sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }$

$\: \:$

$\bf \large \: I_B = 50μA$

$\: \:$

_______________________________

<h3>Collector Current ,</h3>

$\sf \large \: I_C = β * I_B$

$\: \:$

$\sf \large \: I_C = 50 * 50 * 10^{-6}$

$\: \:$

$\bf \large \: I_C = 2.5mA$

$\: \:$

________________________________

<h3> Collector to emitter Voltage : </h3>

$\sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )$

$\: \: \:$

$\sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )$

$\: \:$

$\bf \large \: V_{CE} = 6.5v$

________________________________

<h3>Q - point are :</h3>

$\sf \large \: I_{CEQ} = 2.5mA$

$\: \:$

$\sf \large \: V_{CEQ} = 6.5v$

________________________________

<h3>Q - point located on the DC load line as shown in fig ~</h3><h3 /><h3 /><h3 /><h3 />

________________________________

Hope Helps!:)

$\: \: \: \:$

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drek231 [11]

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