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3 years ago

Gummy gummy bears gummy gummy bears

Engineering

2 answers:

ivanzaharov [21]3 years ago

8 0

I love that song lol

Dima020 [189]3 years ago

8 0

Answer:

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh yeaoooh

Gummy Gummy Gummy Gummy Gummibär

Bai ding ba doli party

Bamm bing ba doli party

Breding ba doli party

Party pop

Bai ding ba doli party

Bamm bing ba doli party

Breding ba doli party

Party pop

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh yeaoooh

Ba Ba Bidubidubi Yum Yum

Three times, you can bite me

Ba Ba Bidubidubi Yum Yum

Three times, you can bite me

Gummy Gummy Gummy Gummy Gummibär

Én vagyok a gumimaci

Gyere már gyere velem

Gyere táncolj már kérlek get get party pop

Én vagyok a gumicummi

Cicci meg a kodile a mambó csoda

Pápé gyere már és táncolj party pop

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Party pop

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Opposition to current flow, restricts or resists current flow

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Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t

Natasha2012 [34]

Answer:

$\dot W_{in} = 273.69\,kW$

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

$\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0$

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

$P = 20\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 251.42\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

Outlet (Subcooled Liquid)

$P = 6000\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 257.502\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

The power input to the pump is computed hereafter:

$\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)$

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The electron concentration in silicon at T = 300 K is given by

puteri [66]

Answer:

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

Explanation:

From the question we are told that:

Temperature of silicon $T=300k$

Electron concentration $n(x)=10^{16}\exp (\frac{-x}{18})$

$\frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})$

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Electron current density $Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2$

Generally the equation for the semiconductor is mathematically given by

$Jn=qb_n\frac{dn}{dx}+nq \mu E$

Therefore

$-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E$

$E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}$

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

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