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3 years ago

How will the proposed study contribute to your career?*(quantity Surveying

Engineering

1 answer:

Flauer [41]3 years ago

5 0

Answer:

PROPOSED STUDY CONTRIBUTES TO YOUR CAREER. Proposed study is essential for career growth. It contributes to our career in a great way. It <u>enhances leadership skills</u> and polish our skills making us more competent. It expand our horizons and opportunities. It gives us better understanding of things. We become able of developing professional relationships with our students as well as development sector which helps in future projects.

Hope this help!:)

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A discrete-time LTI system H has input x[n] and output y[n] related by the linear constant coefficient difference equation y[n]

Molodets [167]

Answer:

See explaination and attachment

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Have the solution as an attachment.

4 0

3 years ago

A 2-lane highway is to be constructed across a 6-ft diameter metal culvert which is oriented perpendicular to the highway center

kvv77 [185]

Answer:

Dude just like forget the highway and drive on the interstate bruh

Explanation:

5 0

4 years ago

Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work

Naily [24]

Answer:

$w_{out}=319.1\frac{BTU}{lbm}$

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

$h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}$

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

$s_2=s_1$

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

$x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965$

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

$h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}$

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

$h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}$

Best regards.

3 0

3 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

3 years ago

(a) The room-temperature electrical conductivity of a semiconductor specimen is 13 (Ω-m)-1. The hole concentration is known to b

Aleksandr-060686 [28]

Answer: a) 0.24E+20 m-3. b) p-type extrinsic.

Explanation:

The current density in a semiconductor is composed by two types of charge carriers: electrons and holes.

This parameter, is proportional to the Electric field within the semiconductor, being the proportionality constant, the electrical conductivity of the material, that takes into account the charge carrier concentrations, and the mobility for each type.

The expression for electrical conductivity is as follows:

σ = q . ne . µe + q . np . µp

Replacing by the given values, and the value of q (charge of an electron), we can get the only unknown that remains, ne , as follows:

ne =( σ – (q . np . µp)) / q µe = (13 (Ω.m)-1 – (1.6E-19) coul(4.0E+20) m-3.0.18) m2/V-s /( (1.6E-19).0.38) coul.m2/V-s

ne = 0.24E+20.

As ne is smaller than np, this means that the semiconductor behaves like a p-type extrinsic one.

7 0

3 years ago

How will the proposed study contribute to your career?*(quantity Surveying​

How will the proposed study contribute to your career?*(quantity Surveying