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3241004551 [841]

3 years ago

10

a poorly tighten terminal is often the cause of a/an ? a) open circuit b) circuit breaker interrupt c)short circuit d) ground fa

ult

1 answer:

Tema [17]3 years ago

7 0

Answer: b) circuit breaker interrupt

Explanation:

Wires connect to outlets through terminals and these terminals need to be tight to ensure that electricity flows with the minimum amount of resistance.

If the terminal is poorly tightened, what happens is that resistance will build up as the appliance is being used and this resistance will make the terminal hotter.

The heat will continue to rise until the circuit breaker trips to interrupt the circuit and protect the appliance.

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If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit

mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

$\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}$

$\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}$

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

$\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}$

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

$\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}$

choose R2, R1 such that it will maintain required ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0

3 years ago

Easy POINTS computer genius help me plz

Norma-Jean [14]

Answer:

dedededededede

Explanation:

4 0

3 years ago

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and

k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

$z1=$ $\frac{\left(150-137\right)}{27.7}$ $=0.4693$

$z2=\frac{\left(201-137\right)}{27.7}=2.3105$

$P(0.4693$

B.)

$z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289$

$\\P(2.9309$

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0

4 years ago

Read 2 more answers

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

max2010maxim [7]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

7 0

3 years ago

The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol

pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = $\frac{\pi}{4}D^2L$

or

V = $\frac{\pi}{4}\frac{D^3}{1.1}$

on substituting the values, we have

245 = $\frac{\pi}{4}\frac{D^3}{1.1}$

or

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

or

L= 6.36 cm

Now,

the compression ratio is given as:

$\textup{compression ratio}=\frac{V+v}{v}$

on substituting the values, we get

$\textup{compression ratio}=\frac{245+27.2}{27.2}$

or

Compression ratio = 10.007

4 0

4 years ago