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3 years ago

What is kirchoff's current law?

Engineering

2 answers:

Digiron [165]3 years ago

5 0

Answer:

answer is 0

Explanation:

hope it helps

sry but I just kinda needed the points :)

Anastaziya [24]3 years ago

4 0

Answer:

The sum of all currents entering a junction is 0

Explanation:

Current cannot disappear, so the currents leaving a junction have to equal the currents entering a junction. If you would give the currents leaving the junction an opposite sign (e.g., negative) it implies that the sum of all these currents is exactly 0.

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A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

3 years ago

Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with

Scilla [17]

Answer:

DIAMETER = 9.797 m

POWER = $\dot W = 28.6 kW$

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

$v =\frac{RT}{P}$

where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

$v = \frac{0.287\times 293}{100}$

v = 0.8409 m^3/ kg

from continuity equation

$A_1 V_1 = A_2 V_2$

$\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2$

$D_2 = D_1 \sqrt{\frac{V_1}{V_2}}$

$D_2 = 8 \times \sqrt{\frac{12}{8}}$

$D_2 = 9.797 m$

mass flow rate is given as

$\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}$

$\dot m = 717.309 kg/s$

the power produced $\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]$

$\dot W = 28.6 kW$

8 0

3 years ago

When we say that communication is

Harlamova29_29 [7]

Answer:

Communication is simply the act of transferring information from one place, person or group to another. Every communication involves (at least) one sender, a message and a recipient.

Explanation:

3 0

3 years ago

You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210

Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0

3 years ago

Ninety-five percent of the acetone vapor in an 85 vol.% air stream is to be absorbed by countercurrent contact with pure water i

Sati [7]

Answer:

Explanation:

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5 0

3 years ago

What is kirchoff's current law?​

What is kirchoff's current law?