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2 years ago

1. In order to minimize hazards, what should you do before starting a job

Engineering

2 answers:

ivanzaharov [21]2 years ago

5 0

Answer:

Explanation:

If you asses the risks you can prepare for the worst

Vilka [71]2 years ago

5 0

C is the correct answer to this question

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Why excess air is required to burn a fuel completely

8_murik_8 [283]

Explanation:

excess air is required to ensure adequate mixing of fuel and air, avoid smoke, minimize sg in Coal burning, and to ensure maximum steam output.

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3 years ago

Read 2 more answers

Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

Fofino [41]

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi ( 1 psi= 6894.76 $\frac{N}{m^2}$ )

we know that $P=\rho\times g\timesh$

where $\rho =Density,g=9.81\frac{m}{s^2}$

h=height.

Given that P=14.769 psi ⇒P= 101828.6 7 $\dfrac{N}{m^2}$

(a) $P=\rho_{w}\times g\times h_{w}$

$\rho_{w}=1000\frac{Kg}{m^3}$

⇒101828.67= $1000\times 9.81\times h_{w}$

$h_{w}$ =10.38 m

So water barometer will read 408.66 in. (1 m=39.37 in)

(b) $P=\rho_{hg}\times g\times h_{hg}$

$\rho_{hg}$ =13600

So 101828.67= $13600\times 9.81\times h_{hg}$

$h_{hg}$ =0.763 m

So mercury barometer will read 30.04 in.

6 0

3 years ago

Ame:<br> 7. A step-down transformer reduces the primary current.<br> True or false

RSB [31]

False

..............

It increases the primary current a step up reduces the primary current

6 0

3 years ago

Read 2 more answers

Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf

docker41 [41]

Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

8 0

3 years ago

An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci

Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0

3 years ago