Storing parts outside doesn’t cause any environmental risks as long as the items are covered.

What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?

nasty-shy [4]

Answer:

$0.14\ lb/in^{3}$

Explanation:

Density is defined as mass ler unit volume, expressed as

$\rho=\frac {m}{v}$

Where m is mass, $\rho$ is density and v is the volume. For a sphere, volume is given as

$v=\frac {4\pi r^{3}}{3}$

Replacing this into the formula of density then

$\rho=\frac {m}{\frac {4\pi r^{3}}{3}}$

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

$\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}$

4 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

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

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

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

7 0

2 years ago

A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying

Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius, $r_{2}$ = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, $r_{1}$ = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

τmax $= \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )$

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

3 0

3 years ago

A group of n Ghostbusters is battling n ghosts. Each Ghostbuster carries a proton pack, which shoots a stream at a ghost, eradic

babunello [35]

Answer:

Using the above algorithm matches one pair of Ghostbuster and Ghost. On each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are the same, so use the algorithm recursively on each side of the line to find pairings. The worst case is when, after each iteration, one side of the line contains no Ghostbusters or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P( $n^{2} lg n$ )- time algorithm.

4 0

3 years ago

Based on the graphs of stress-strain from the V-MSE site, how would you characterize the general differences between polymers an

Pepsi [2]

Answer:

Option A

Explanation:

Alloys are metal compounds with two or more metals or non metals to create new compounds that exhibit superior structural properties. Alloys have high level of hardness that resists deformation thereby making it less ductile compared to polymers. This is due to the varying difference in the chemical and physical characteristics of the constituent metals in the alloy.

6 0

3 years ago