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3 years ago

Alright, who’s dirty minded cause i am :)

Engineering

2 answers:

densk [106]3 years ago

8 0

Answer:

me;)

Explanation:

wel3 years ago

5 0

Answer:

I am lol

Explanation:

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g Calculate a better value for the convection coefficient using resources from heat transfer. Assume forced convection of the ai

ollegr [7]

Answer:results for h should on the same order of magnitude of the value I provided. If it isn't, check your units. Update to you calculated values for h and resolve your model.

Explanation:

4 0

3 years ago

State three means of operating a power tool

Shtirlitz [24]

Power tools are usually run on one of the three types of power: Compressed air, electricity, or combustion

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2 years ago

When charging an R-410A system that uses a water-cooled condenser, you must first charge with vapor to a pressure of at least __

kirza4 [7]

Answer:

35psig

Explanation:

Pressure is a force that acts on a surface area. The psig measure the pressure pounds per square inch gauge. It measure the difference in pressure between supply tank and outside air. This pressure is relative to atmospheric pressure. When charging an R-410A system, the vapor pressure should be at least 35 psig before switching to liquid charging.

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4 years ago

Explain Laser Doppler Vibrometer in details.

SOVA2 [1]

Answer:

A laser Doppler vibrometer (LDV) is a science tool used to evaluate a surface's non-contact vibration

Explanation:

laser Doppler vibrometter (LDV) is a science tool used to evaluate a surface's vibration without in contact to the surface. The LDV laser beam is directed to the surface of significance & the wave amplitude, frequency is obtained from the reflected laser beam frequency's Doppler change due to ground movement. it identifies the Doppler effect or shift of laser light reflected from the sample object

6 0

4 years ago

Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)

Gre4nikov [31]

Answer:

Explanation:

Given that:

$y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)$

where;

$g'(0) = \dfrac{1}{m}$ and $mg"+kg = 0$

$\text{Using Leibniz Formula to prove the above equation:}$

$\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt$

So, $y = \int ^t_0 g' (t-s) f(s) \ ds$

$\text{By differentiation with respect to t;}$

$y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds$

$y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)$

$Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\$

$Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}$

7 0

3 years ago