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3 years ago

What are the two tests described here with abbreviations?

Engineering

1 answer:

Alex73 [517]3 years ago

3 0

Answer:

Scott was in a motorcycle accident and has injuries to his chest and his skull. The doctor has ordered Scott to have an ECG, which stands for ____, to check his heart rhythm

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Design a filter that has infinite DC gain, a gain of one from 1Hz to 100 Hz and filters (1storder) any signals above 100 Hz.a) S

EastWind [94]

Answer:

Attached below are the sketches

answer :

c) G(s) = 100 / ( s + 100 )

d) y'(t) + 100Y(s) = 100 X(s)

e) g(t) = e^-100t u(t)

Explanation:

a) Sketch the bode plot

The filter here is a low pass filter

b) Sketch the s-plane

attached below. pole ( s ) is at 100

c) write the transfer function of the filter

Transfer function ; G(s) = 100 / ( s + 100 )

d) write the differential equation

Y(s) / X(s) = 100 / s + 100

Y(s) [ s + 100 ] = 100 X(s)

= sY(s) + 100Y = 100 X(s)

∴ differential equation = y'(t) + 100Y(s) = 100 X(s)

e) write out the unforced transient response

g(t) = e^-100t u(t)

f) write out the frequency response

attached below

4 0

3 years ago

For a metal that has an electrical conductivity of 6.1 × 107 (Ω∙m)–1, what is the resistance of a wire that is 4.3 mm in diamete

kotegsom [21]

Answer: (C) 9.14 . 10⁻³ Ω

Explanation:

The resistance of a resistor, is proportional to his length and inversely proportional to his area, being the proportionality constant a property of the material, called resistivity.

The resistivity is defined as the inverse of the electrical conductivity, which depends on the number of charge carriers and the mobility of these carriers, which is different for each material.

So, we can calculate the resistance as follows:

R = 1/σ . L / A, where:

σ = electrical conductivity, l= length of the wire , A = wire cross-section (assumed circular).

Replacing by the values, we can calculate R as follows:

R = 1/6.1. 10⁷ (Ω.m) . 8.1 m. / π (0.0043)² m / 4 = 9.14 . 10⁻³ Ω

7 0

3 years ago

A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=

Dafna11 [192]

Answer:

$v_{o} = 22.703\,\frac{m}{s}$ $\left(50.795\,\frac{m}{s}\right)$

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

$\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}$

Where:

$a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g$

$a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$a_{1} = -3.040\,\frac{m}{s^{2}}$

Likewise, the deceleration of the car on the unpaved shoulder is:

$a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g$

$a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$a_{2} = -2.549\,\frac{m}{s^{2}}$

The speed just before the car entered the unpaved shoulder is:

$v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}$

$v_{o} = 18.175\,\frac{m}{s}$

And, the speed just before the pavement skid was begun is:

$v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}$

$v_{o} = 22.703\,\frac{m}{s}$ $\left(50.795\,\frac{m}{s}\right)$

8 0

3 years ago

Select the correct answer.

andre [41]

Answer:

A. energy transformations

Explanation:

8 0

2 years ago

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidi

padilas [110]

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

5 0

3 years ago