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Ne4ueva [31]
3 years ago
10

I accidently peed my pants help me change me pls

Engineering
2 answers:
nataly862011 [7]3 years ago
3 0

Answer: *changed*

Explanation: Because you peed

svetoff [14.1K]3 years ago
3 0

Answer:ok but I’m gay so if you a girl sorry I can’t get heat now.

Explanation:

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What is engineering
kirill [66]

Answer:

Engineering is the use of scientific principles to design and build machines, structures, and other items, including bridges, tunnels, roads, vehicles, and buildings. In other words, Engineering is the use of science and math to design or make things.

Explanation:

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Which of the following are some of the problems found in city streets?
storchak [24]

Answer: drugs and rushing cars

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rushing cars or reckless drivers cut curb fast and potentially someone can get hurt they are speeding and not worrying about other people lives at stake

4 0
3 years ago
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
An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh
Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= \frac{M_{mm}Y }{I}  = \frac{6.48 * 10^6 * 2.375}{53.54} =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= \frac{M_{mm}Y }{I} = \frac{6.48 *10^6*(8.5-2.375)}{53.54} = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = \frac{wL}{2} = \frac{1000*6*12}{2} = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - \frac{0.5}{2} ) * \frac{(2.375 - (\frac{0.5}{2} ))}{2}

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = \frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}

= 1.143 *10^5

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hmm tdyufrjhfjjbdjjkljgjjddtu

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