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klio [65]
3 years ago
14

Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______

Engineering
1 answer:
Alina [70]3 years ago
5 0

Answer:

Multiplying impulse response by t  ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

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C

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Why is tubing sometimes coiled when installed in a car or vechile
andriy [413]

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Coiled tubing is often used to carry out operations similar to wire lining.

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The simple majority decision rule may generate results that are
Arisa [49]

Answer:

Efficient when the marginal benefits of project = marginal costs of project.

Explanation:

Majority Decision Rule:

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5 0
3 years ago
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h
gtnhenbr [62]

Answer: 5.36×10-3kg/h

Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.

Explanation:using the formula

M =JAt = -DAt×Dc/Dx

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6 0
3 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
3 years ago
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