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Ronch [10]
3 years ago
6

A _______ contact allows current to flow when the switch's operator is not activated.?

Engineering
1 answer:
Anna35 [415]3 years ago
3 0

Answer:

Normally closed (NC)

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

A switch can be defined as an electronic component or device that is typically used to prevent or allow the flow of electric current (electrons) from one point to another in an electrical circuit.

There are two logical forms used for designing a switch and these includes;

I. Normally opened (NO).

II. Normally closed (NC).

A Normally closed (NC) contact allows current to flow when the switch's operator is not activated.

Basically, a Normally closed (NC) contact would always remained closed unless a certain (specific) condition is met. Thus, when the contacts are moved, the circuit becomes open and thus, prevents the flow of electric current in the circuit.

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The federal highway administration reports nearly
levacccp [35]

Explanation:

The federal highway administration reports nearly 800 work zone fatalities per year.

8 0
3 years ago
Consider a 6-bit cyclic redundancy check (CRC) generator, G = 100101, and suppose that D = 1000100100. 1. What is the value of R
dmitriy555 [2]

Answer:

The value of R is 10101

Explanation:

As per the given data

D = 1000100100

G = 100101

Redundant bit = 6-bits - 1-bit = 5-bits

No add fice zero to D

D = 100010010000000

Now calculate R as follow

R = D / G

R = 100010010000000 / 100101

R = 10101

Workings are attached with this question

3 0
3 years ago
A 14-lb crate is pulled up a frictionless 40° ramp with an initial velocity of v1=0.4 ft/s. It is pulled 0.3 ft from location #1
Morgarella [4.7K]

Answer:

3.25 ft/s

Explanation:

The crate is of =14-lb=m₁

The angle of inclination is = 40°=Ф

The initial velocity = 0.4 ft/s= v₁

Distance the crate will move is= 0.3 ft =d

The load pulling downwards is = 36 lb= m₂

Acceleration of the pulley, a= m₂g - m₁gsinФ / m₁+m₂ where g= 32.17 ft/s^2

a= 36*32.17 - 14*32.17*sin 40° / 14+36

a=17.37 ft/s^2

Apply the formula for final velocity

V₂²=V₁²+2ad

V₂²=0.4²+ 2*17.37*0.3

V₂²=10.582

V₂ =√10.582 = 3.25 ft/s

6 0
3 years ago
4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and
Alika [10]

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

 

Since the area is  

Time (min)           Runoff (cm)       Volume of runoff m³

0                   0                                     0

10                  4                                     1000000 m³

20                 2.5                                  625000 m³

30                 2                                      500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

For the 1st  10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

 

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

 

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

6 0
3 years ago
Nowadays power supply on board ship is determine low and high voltage. From what range low & high voltage is being determine
zlopas [31]

Any Voltage used on board a ship if less than 1kV (1000 V) then it is called as LV (Low Voltage) system and any voltage above 1kV is termed as High Voltage. Typical Marine HV systems operate usually at 3.3kV or 6.6kV.

Explanation:

trust

6 0
1 year ago
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