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Ugo [173]
3 years ago
8

In the layout of a printed circuit board for an electronic product, 12 different locations can accommodate chips.

Engineering
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

a) 244,140,625 different ways

b) 390,625 different ways

Explanation:

a) If there are 5 ways to place a chip on each location, and there are 12 locations overall, we have:

5^12 ways of placing them

This would mean a total of 244,140,625 different ways

b) If five chips are of the same type, we can first find how many ways we can place chips on the remaining 7 locations:

5^7 = 78,125

Next we can multiply this by the number of ways the next 5 chips could be the same:

78,125 * 5 = 390,625 different ways

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10. True or False? A disruptive technology<br> radically changes the way people live and<br> work.
Anettt [7]

Answer:

<u>True</u>

Explanation:

According to Investopedia.com, "Disruptive technology is an innovation that significantly alters the way that consumers, industries, or businesses operate".

So yes, a disruptive does radically change the way people live and work.

5 0
3 years ago
Read 2 more answers
The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra
Lina20 [59]

Answer:32.4m/s^2

Explanation:

Given data

radius\left ( r\right )=0.4m

Intial angular velocity\left ( \omega_0\right )=4rad/s

angular acceleration\left ( \alpha\right )=5rad/s^2

angular velocity after 1 sec

\omega=\omega_0+\alpha\times\t

\omega=4+5\left ( 1\right )

\omega=9rad/s

Velocity of point on the outer surface of disc\left ( v\right )=\omega_0\timesr

v=9\times0.4 m/s=3.6m/s

Normal component of acceleration\left ( a_c\right )=\frac{v^2}{r}

a_c=\frac{3.6\times3.6}{0.4}=32.4m/s^2

3 0
4 years ago
For a metal casting with a cylindrical riser, what is the best (optimal) aspect ratio (aspect ratio is the ratio of the riser's
DochEvi [55]

Answer:

Optimal aspect ratio = 1

Explanation:

From Chvorinov's Rule,

T = c [\frac{volume}{surface are}]^n

Where

T = solidification time

c = mold constant

Volume = \frac{pi D^2 h}{4}

Surface areaA= pi Dh[\frac{2pi D^2}{4}

For the riser to have the longest solidification time:

We have: D = h

V = \frac{pi D^3}{4}

A = pi Dh[\frac{2pi D^2}{4}] = 1.5D^2 pi

Therefore,

\frac{V}{A} = \frac{\frac{pi D^3}{4}}{1.5D^2 pi}

\frac{V}{A} = \frac{D}{6}

To have to longest solidification time, D = h

Therefore the best aspect ratio would be,

\frac{D}{h} = 1

3 0
3 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe
LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is    = 1.751 \times 10^{-5} \% per hr

Explanation:

we know Arrhenius expression is given as

\dot \epsilon =Ce^{\frac{-Q}{RT}

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is\dot \epsilon = 5.5\times 10^{-2}% per hr

At 800 degree C  creep rate is\dot \epsilon = 1% per hr

activation energy for creep is \frac{\epsilon_{800}}{\epsilon_{700}} = = \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}

\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}

\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

C = \epsilon e^{\frac{Q}{RT}}

    =- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr

\epsilon_{500} = C e^{\frac{Q}{RT}}

                         = 1.804 \times 10^{12}  e{\frac{251758}{8.314(500+273}}

                         = 1.751 \times 10^{-5} \% per hr

4 0
3 years ago
Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at
andriy [413]

Answer:

A) v_2 = 2016.80 ft/s

B) \Delta s = 0.006 Btu/lbm R  

Explanation:

Given data:

P-1 = 100 lbf/in^2

T_1 = 500 degree f

V_1 = 100 ft/s

P_2 = 40 lbf/inc^2

effeciency = 80%

from steady flow enerfy equation

h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

H_1 = 1278.8 Btu/lbm

s_1 = 1.708 Btu/lbm -R

for P1 = 40 lbf/in^2

H_1 = 1193.5 Btu/lbm

s_1 = 1.708 Btu/lbm -R

exit enthalapy h_2

\eta = \frac{h_1 - h'_2}{ h_1 - h_2}

0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm

from above equation

1278.8 \times 25037 + \frac{100^2}{2} = 1197.77   \times 25037 + \frac{v_2^2}{2}                   [1 Btu/lbm = 25037 ft^2/s^2]

v_2 = 2016.80 ft/s

b) amount of entropy

\Delta s = s_2 - s_1

s_1 = 1.708 Btu/lbm -R

at h_2 = 1197.77 Btu/lbm [\tex]  and [tex]P_2 = 40 lbf/in^2

s_2 is 1.714 Btu/lbm -R

\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R

6 0
3 years ago
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