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2 years ago

Kirchoff's Law states that, by the time current has returned to its source, all

Engineering

2 answers:

Alchen [17]2 years ago

6 0

Kirchoff's law states that by the time current has reached to its source all voltage must be used.

<u>Explanation:</u>

Kirchoff's voltage law states that the algebraic sum of all the voltages across a closed loop is zero. As the circuit is a closed conducting path, no energy is lost. ices All the voltage that is supplied from the voltage source is dropped at the devices in the circuit.

The potential difference maintained by the voltage source enables the flow of current across the circuit. As the supplied voltage drops at each elements of the circuit, all voltage is used by the time current has returned to the surface.

Alex17521 [72]2 years ago

4 0

Kirchoff's Law Kirchoff's Law states that, by the time current has returned to its source is explained in the following.

Explanation:

Kirchhoff's Current Law (KCL) is Kirchhoff's first law that deals with the conservation of charge entering and leaving a junction. ... In other words the algebraic sum of ALL the currents entering and leaving a junction must be equal to zero.

Kirchoff's laws apply for a given instant in time. So the voltages at a given moment around a loop will all sum to zero, or currents in a node sum to zero if you look at the instantaneous voltage and current. But they will be out of phase.

Kirchhoff Voltage Law states that ''The algebraic sum of all voltages (source voltage and voltage drops) is equal to zero around a close path''. This is called KVL ( Kirchhoff Voltage Law) equation. The source voltage is equal to the sum of all voltage drops.

Kirchhoff's Voltage Law (KVL) is Kirchhoff's second law that deals with the conservation of energy around a closed circuit path.

Kirchhoff's laws can be used to determine the values of unknown values like current, Voltage in the circuit. These laws can be applied on any circuit (with some limitation), and useful to find the unknown values in complex circuits and networks.

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Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

Energy = 1.4579 eV

Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

Wavenumber is the reciprocal of wavelength.

So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

5 0

2 years ago

The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm

Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0

2 years ago

A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1

ella [17]

Answer:

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

first we have to determine the total TMU = ∑ TMU = 86.4 TMU

note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

7 0

1 year ago

What’s the difference between quality and quantity

Lesechka [4]

Quality is how good something is. Like how good the material is. Like the quality of the IPhone X is better than the quality of the iPhone 6s + because it has more features and can sustain better hold
Quantity is how much. Like the quantity of the m&ms were split evenly.
I dont know exactly if that makes sense but there ya go
YW!

4 0

2 years ago

Read 2 more answers

What is the ANSI B paper size also know as?

krek1111 [17]

Answer:

ANSI A sized paper is commonly referred to as Letter and ANSI B as Ledger or Tabloid.

Explanation:

7 0

2 years ago