Easy POINTS computer genius help me plz

Study the graphs below what type of trend and pattern in observed .

hichkok12 [17]

Answer:

negative and linear

Explanation:

its a straight line going down

8 0

3 years ago

Read 2 more answers

What system transfers the power created by the engine to the

Juli2301 [7.4K]

Pretty sure the answer is A

8 0

4 years ago

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In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium

Nimfa-mama [501]

Answer:

The diffusivity is given as $8.064\times 10^{-16} m^2/s$

Explanation:

From the given data as in the attached question found via the search (because the values were not clear in this one)

$D_o=2.0\times10^{-4}m^2/s\\Q_d=240.6 kJ/mol=2.406 \times 10^5 J/mol\\Gas Constant=R=8.314 J/mol K\\Temperature =830 C =830+273 =1103 K$

So the Diffusion coefficient is given as

$D=D_oe^{\frac{-Q_d}{RT}}\\D=(2.0\times 10^{-4})e^{\frac{-2.406\times 10^5}{8.314\times 1103}}\\D=8.0640\times 10^{-16} m^2/s$

So the diffusivity is given as $8.064\times 10^{-16} m^2/s$

4 0

4 years ago

Cryogenic liquid storage. Liquid oxygen is stored in a thin-walled spherical container, 96 cm in diameter, which is further encl

Anvisha [2.4K]

Answer:

The answer is "26.55 V"

Explanation:

Given values:

$d_i= 0.96m\\d_o= 1m\\\epsilon = 0.05\\T_0= 280k\\T_i= 95k\\$

For Answer (a) please find the attachment.

Answer (b):

$q_{i-0}= \frac{\sigma (T_{0}^4)-(T_{i}^4)}{\frac{1-\epsilon i }{\epsilon_{i} A_{i}}+ \frac{1 }{\ f_{i o} A_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0} A_{0}}}$

$f_{i0}= 1 \ it \ is \ fully \ inside \ the \ large \ sphero \\$

$q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} - 1+ 1 +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times \frac{A_i}{A_0}}\\\\q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{d_i}{d_0})^2}\\\\q_{i-0}= \frac{\sigma (\pi d^2_i) (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{r_i}{r_0})^2}\\\\$

$q_{i-0}= \frac{5.67 \times 10^{-8} \times 3.14 \times 9.62 \times 9.62 \times (280^4-94^4)}{\frac{1 }{0.05} +\frac{1-0.05}{0.05} \times (\frac{0.96}{1})^2}\\\\\ After \ solve the \ equation \ the \ answer \ is:\\\\q_{i-0} = 26.55 \ V$