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3 years ago

THIS IS NOT AN ACADEMIC QUESTION, but who was the bitter of 1987 in FNAF?

Engineering

2 answers:

ivann1987 [24]3 years ago

4 0

Answer:

Golden Freddy

Explanation:

Fredbear did the bit, but after about 15 years, he became Golden Freddy

Genrish500 [490]3 years ago

3 0

Wasn’t it Fred bear?

You might be interested in

A(n)_____ is a device that provides the power and motion to manipulate the moving parts of a valve or damper used to control flu

Lesechka [4]

Answer:

Out of the four options provided

option A. actuator

is correct

Explanation:

An actuator is the only device out of the four mentioned devices that provides power and ensures the motion in it in order to manipulate the movement of the moving parts of the damper or a valve used whereas others like ratio regulator are used to regulate air or gas ratio and none mof the 3 remaining options serves the purpose

5 0

3 years ago

A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is

Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

6 0

3 years ago

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length

masya89 [10]

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 \times 10^{-5} kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s$

Time period $= \frac{210}{5.04\times 10^{-3}} = 41654.08 s$

3 0

3 years ago

Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new

Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

$C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$ where i=1 to N-1 and N=6

$C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$

A area of the polygon can be found by the following formula $A=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i})$ where i=1 to N-1

$A=\frac{1}{2}[ (x_{1} y_{2} -x_{2} y_{1})+ (x_{2} y_{3} -x_{3} y_{2})+(x_{3} y_{4} -x_{4} y_{3})+(x_{4} y_{5} -x_{5} y_{4})+(x_{5} y_{6} -x_{6} y_{5})]$

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

$C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{x} =15.36$

For Cy

$C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(y_{2}+y_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(y_{3}+y_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(y_{4}+y_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(y_{5}+y_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{y} =22.55$

So coordinates for the fire station should be (15.36,22.55)

5 0

3 years ago

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = $\frac{q^2Nd^2*Xn^2}{6Eo} * f$

also note that ; Pactive ∝ Nd2 (

tD = K . $\frac{Vdd}{(Vdd - Vt )^2}$ since K = constant

Hence : Nd ∝ rt

5 0

3 years ago