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2 years ago

Can someone draw a electric circuit with electrical symbols for this

Engineering

2 answers:

erastovalidia [21]2 years ago

4 0

Answer:

i cant draw it

i am sry

Explanation:

Ksenya-84 [330]2 years ago

3 0

Answer:

Yes, there can be a symbol drawn for this.

Explanation:

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At a festival, spherical balloons with a radius of 140.cm are to be inflated with hot air and released. The air at the festival

Tpy6a [65]

Answer:

find attached

Explanation:

5 0

3 years ago

The diffusion coefficients for species A in metal B are given at two temperatures:

Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

$D = D_0e^{\frac{-Q_d}{RT} }$

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

$R = 8.314\ J/mol*K$

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

$ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}$

and

$ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}$

You might notice that these equations have the form of

$d=y-ax$

You can solve this equation system easily using calculator, and you will eventually get

$D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol$

After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation

$6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}$

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0

2 years ago

How much metal can be removed from a cracked drum to restore surface

Alisiya [41]

Answer:sc

Explanation:

7 0

2 years ago

Read 2 more answers

A sleeve made of SAE 4150 annealed steel has a nominal inside diameter of 3.0 inches and an outside diameter of 4.0 inches. It i

irga5000 [103]

Answer:

Class of fit:

Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).

Here minimum shaft diameter will be greater than the maximum hole diameter.

Medium Drive Force Fits are FN 2 Fits.

As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

Max Shaft Diameter: 3.0029

Min Shaft Diameter: 3.0022

Max Hole Diameter:3.0012

Min Hole Diameter: 3.0000

Max Interference: 0.0029

Min Interference: 0.0010

Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.

Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

Explanation:

4 0

3 years ago

Read 2 more answers

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

2 years ago