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Gnom [1K]
4 years ago
12

P(5)=(0.11 x14)^5/5!e-^0.11-14

Engineering
1 answer:
seraphim [82]4 years ago
7 0

Answer:

The value of P(5)=0.09264

Explanation:

The equation given is;

P(5)=\frac{(0.11*14)^5}{5!e^-0.11-14} \\\\\\P(5)=(1.54)^5/120*2.7183^{-0.11} -14\\\\\\P(5)=8.662/120*0.8958-14\\\\\\P(5)=8.662/107.5-14\\\\P(5)=8.662/93.5\\\\P(5)=0.09264

The value of P(5)=0.09264

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What steel type and ASTM designation is preferred for W-shapes?
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Answer:

The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.

Explanation:

The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.

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4 years ago
What is tear test? What do we understand from this test? How?
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Explanation:

The tear test determines the force required by a material to undergo complete failure when there is already a crack or tear present in it.

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4 years ago
Cell phones require powerful batteries in orde to work effectively. Which activity is best described as an engineering endeavor
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8 0
3 years ago
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})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

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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3 years ago
Explain why the following scenario fails to meet the definition of a project description.
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Answer:

The youth hockey training facility

Explanation:

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