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marissa [1.9K]
3 years ago
15

There are gauges available for checking panel flushness. TRUE OR FALSE?

Engineering
2 answers:
jolli1 [7]3 years ago
7 0

Answer: true

Explanation:

because flushness means your dead   so you are dead now

Anit [1.1K]3 years ago
4 0
There are gauges available for checking panel flushness. to allow for front-to-back movement. to allow for side-to-side movement. for height adjustments. So yes it is true. Hope this helps!
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g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus
iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

\sigma = \frac{F}{A}\\\\

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2

Therefore,

\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips

Now, we determine the strain:

strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon =  7.5\ x\ 10^{-4}

Now, the modulus of elasticity (E) is given as:

E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}

<u>E = 8.83 kips</u>

7 0
3 years ago
QUESTION
Ymorist [56]
Torch body if I’m wrong I’m really sorry that’s what I got
3 0
3 years ago
If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo
Thepotemich [5.8K]

Answer:

The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)} and considering that E(s)=D(s)-G_c(s)C(s) then

\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)

\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)

\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}

E(s)=\frac {D(s)}{1+G_c(s)G(s)}

For ramp disturbance d(t)=at

D(s)=\frac {a}{s^{2}} therefore, the steady state error is given by

e(\infty)= \lim_{s \to 0} s E(s)

e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]

Whenever G_c(s) has a double intergrator, the error e(\infty) becomes zero

3 0
3 years ago
How does the "E" in STEM work with the other letters
KIM [24]

Answer:

if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society

4 0
4 years ago
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