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Leto [7]
2 years ago
5

What is voltage drop?

Engineering
1 answer:
lord [1]2 years ago
6 0

Answer:

Explanation:

It is the voltage a voltmeter would read when connected across something that has resistance.

___________0_________O______O_______

|                                                                               |

|                                              |                                |

|_____________________|  |_____________ |

                                              |

The diagram above is supposed to represent 3 lightbulbs connected in series. The vertical lines in the middle are supposed to be a battery which powers the three light bulbs. If you put a voltmeter across one of the lightbulbs, it will read a voltage that is 1/3 of the voltage of the battery.

Answer

That reading you get across the one light bulb is The Voltage Drop.

                                             

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Explain three examples of workshop
stealth61 [152]

Answer:

Invitational Workshop

An invitational workshop is what many of us know. It’s what Lucy Calkins has made famous through the Reading and Writing Workshop. In the invitational workshop, the instructor usually hosts a minilesson. This minilesson is intended to meet the needs of the majority of children in the classroom. Afterward, the children are “invited” to employ the skills or strategy for the minilesson during workshop time, where students work independently or in small groups

Explanation:

7 0
2 years ago
Read 2 more answers
PLEASE HELP, THANK YOU
Elena L [17]

Answer:

6,3,2,5,1,4 because they jst are

Explanation:

3 0
3 years ago
Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96
Viktor [21]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

5 0
3 years ago
Can be used to eliminate rubbing friction of wheel touching frame. 1.Traction 2.Thrust washer
Vilka [71]

Answer:

thrust washer

can be used to eliminate rubbing friction of wheel touching frame

5 0
3 years ago
Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158
ahrayia [7]

Answer:

v_2 = 160.23 m/s

T_2 = 475.797 k

Explanation:

given data:

Diameter =d_1 = 200mm

t_1 =195 degree

p_1 =500 kPa

v_1 = 100m/s

p_2 = 85kPa

d_2 = 158mm

from continuity equation

A_1v_1 = A_2v_2

v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}

v_2 = \frac{d_2v_1}{d_2^2}

v_2 = [\frac{d_1}{d_2}]^2 v_1

      = [\frac{0.200}{0.158}]^2 \times 100

v_2 = 160.23 m/s

by energy flow equation

h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w

z_1 =z_2 and q =0, w =0 for nozzle

therefore we have

h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}

dh = \frac{1}{2} (v_1^2 -v_2^2)

but we know dh = Cp dt

hence our equation become

Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)

Cp (T_2 -T_1) = 7836.94

(T_2 -T_1) = \frac{7836.94}{1.005*10^3}

(T_2 -T_1) = 7.797

T_2 = 7.797 +468 = 475.797 k

8 0
3 years ago
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