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3 years ago

Who invented a control unit for an artificial heart? ements ante

Engineering

1 answer:

disa [49]3 years ago

6 0

Otis Boykin is the answer.

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase

ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ( $\% V_{Gr}$ ) is determined by the following expression:

$\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%$

Where:

$V_{Gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{Fe}$ - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

$V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}$

$V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}$

Where:

$m_{Gr}$ , $m_{Fe}$ - Masses of the graphite and ferrite phases, measured in grams.

$\rho_{Gr}$ , $\rho_{Fe}$ - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

$\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%$

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

$m_{Gr} = \frac{2.5}{100}\times (100\,g)$

$m_{Gr} = 2.5\,g$

$m_{Fe} = 100\,g - 2.5\,g$

$m_{Fe} = 97.5\,g$

If $m_{Gr} = 2.5\,g$ , $m_{Fe} = 97.5\,g$ , $\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}$ and $\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}$ , the volume percent of graphite is:

$\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%$

$\% V_{Gr} = 91.906\,\%$

The volume percent of graphite is 91.906 per cent.

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3 years ago

What can be used to relieve stress in a weld.

satela [25.4K]

Yes , of course you can

3 0

3 years ago

Read 2 more answers

The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

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3 years ago

we want to make a schottky diode on one surface of an n-type semiconductor, and an ohmic contact on the other side. the electron

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Answer:

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4 years ago

A BS of 5.43 ft is taken on a level rod at a 120-ft distance, and a FS of 8.76 ft is taken on the rod held 1,100 feet away.(a) W

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Answer:

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Explanation:

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4 years ago