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Luba_88 [7]
2 years ago
11

What car is this? I thinks its a nissan 240sx but i dont know

Engineering
1 answer:
Pie2 years ago
4 0

Answer:

your right I had to look a little bit but you are correct

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Pressure sensor rated at 0-500 psi and has an output of 0-10 Volts DC (10 V correspond to
scoray [572]

Answer:

Explanation:

Sorry I'm new and I need points ty

5 0
3 years ago
List and explain the major features that the following building designs must have to relate directly to their functions.
Lunna [17]

Answer:

size, design

Explanation:

''.''

7 0
3 years ago
A. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy f
NISA [10]

Answer:

a. Fraction of Atom = 2.41E-5 when T = 600K

b. Fraction of Atom = 5.03E-10 when T = 298K

Explanation:

a.

Given

T = Temperature = 600K

Qv = Energy for formation = 0.55eV/atom

To calculate the fraction of atom sites, we make use of the following formula

Nv/N = exp(-Qv/kT)

Where k = Boltzmann Constant = 8.62E-5eV/K

Nv/N = exp(-0.55/(8.62E-5 * 600))

Nv/N = 0.000024078672493307

Nv/N = 2.41E-5

b. When T = 298K

Nv/N = exp(-0.55/(8.62E-5 * 298))

Nv/N = 5.026591237904E−10

Nv/N = 5.03E-10 ----- Approximated

6 0
4 years ago
In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
gladu [14]

Answer:

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}

Where nCx is the combinatorial number

nCx=\frac{n!}{x!(n-x)!}

Let's calculate f(x) :

f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064

f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288

f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432

f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216

For the cumulative distribution function that we are looking for :

P(X\leq x)=F(x)

F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1

F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1

The cumulative distribution function for X is :

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

5 0
3 years ago
2.11 Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface
alexandr402 [8]

Answer:

The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.

The cost associated with the heat loss through the windows for an 8-hour flight is:

For aerogel windows: $17.472 (most efficient)

For polycarbonate windows: $262.08

For soda-lime glass windows: $1,238.016 (least efficient)

Explanation:

To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:

\frac{dQ}{dt}=\dot Q =-kS\nabla \vec{T}

In this case:

\dot Q =k\frac{S}{L} \Delta T

If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).

To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).

To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):

Cost=C_{hc}\cdot \dot Q \cdot t \cdot n=1\frac{\$}{Kwh} \cdot \dot Q \cdot 8h \cdot 130

6 0
3 years ago
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