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Semenov [28]
3 years ago
7

I need unseen passage

Engineering
2 answers:
monitta3 years ago
8 0

Answer:

this unseen passage__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Semmy [17]3 years ago
6 0

Answer:

An unseen passage is invisible on ground, but might be underground.

Explanation:

I'm sorry but if your question was clear, I would be able to answer it.

Hope this helps!

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I don’t like points so here’s 50.<br><br> What are the following cars?
Alex Ar [27]

Answer:

first one is a oldsmobile cutlasse  2nd is a 65- 69 mustang 3rd is another mustang and the fourth is a dodge

5 0
3 years ago
A low-altitude meteorological research balloon, temperature sensor, and radio transmitter together weigh 2.5 lb. When inflated w
gogolik [260]

Answer:

12 mins

Explanation:

The summation of the forces in vertical direction

= Fb - Fd - w = 0 ∴ Fd = Fb - w  ----- ( 1 )

Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )

Pair = air density , Vballoon = volume of balloon

Vballoon = \frac{\pi D^3}{6} ,  where D = 4  ∴  Vballoon = 33.51 ft^3

g = 32.2 ft/s^2

From property tables

Pair = 2.33 * 10^-3 slug/ft^3

μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s

<em>Insert values into equation 2 </em>

Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 )  = 2.514 Ib

∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )

Assuming that flow is Laminar  and  RE < 1

Re = (Pair * vd) / μair   -------- ( 3 )

where:  Pair = 2.33 * 10^-3 slug/ft^3 ,  vd = ( 987 * 4 ) ft^2/s ,   μair = 3.8 * 10^-7 slug/ft.s

Insert values into equation 3

Re = 2.4 * 10^7 (  this means that the assumption above is wrong )

Hence we will use drag force law

Assume Cd = 0.5

Express  Fd using the relation below

Fd = 1/2* Cd * Pair * AV^2

therefore V = 1.39 ft/s

Recalculate  Reynold's number using  v = 1.39 ft/s

Re = 34091

from the figure Cd ≈ 0.5 at Re = 34091  

<u>Finally calculate the rise time ( time taken to reach an altitude of 1000 ft ) </u>

t = h/v

  = 1000 / 1.39 = 719 seconds ≈ 12 mins

3 0
3 years ago
A) A cross-section of a solid circular rod is subject to a torque of T = 3.5 kNâ‹…m. If the diameter of the rod is D = 5 cm, wha
Alisiya [41]

Answer:

\tau_{max}  = 142.6 MPa

T = 1536.8 N m

Explanation:

Given data:

Torque = 3.5 k N m = 3.5*10^3 N.m

Diameter D = 5 cm = 0.05 m

a) from torsional equation we have

\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}

\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}

solving for \tau_{max}

\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}

\tau_{max}  = 142.6 MPa

B)

\tau = 37 MPa = 37 \times  10^6 Pa

D_i = 4.5 cm = 0.045 m

D_o = 6.5 cm = 0.065 m

\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}

\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}

T = 1536.8 N m

7 0
4 years ago
How may a Professional Engineer provide notice of licensure to clients?
Vsevolod [243]

Find full question attached

Answer:

(b) By including a statement that he or she is licensed by the Board for Professional Engineers and Land Surveyors immediately above the signature line in at least 12 point type on all contracts for services

Explanation:

A PE(professional engineer) licensee must show that he is licensed in order to show and ensure public safety as he is qualified for the job he is handling. The California regulations on professional engineers holds that all professional engineers must be licensed by the board of professional engineers and Land surveyors in order to operate legally as an engineer. The engineer may show licensure through the following options:

The engineer might provide statement to each client to show he is licensed which would then be signed by the client

The engineer may choose to post a wall certificate in his work premises to show he is licensed

The engineer may choose to include a statement of license in a letterhead or contract document which must be above the client's signature line and not less than 12 point type

7 0
3 years ago
In a foundry, metal castings are cooled by quenching in an oil bath. Typically, a casting weighting 20 kg and at a temperature o
Irina18 [472]

Answer:

4.18 J/KgK

Explanation:

Equilibrium point is reached when

m₁*c₁(T₁-T) = m₂*c₂(T -t₂)

m1 = 20

c1 = 0.5

T1 = 450

m2 = 150 kg

c2 = 2.6

T2 = 50

putting these values into the formula

20*0.5 (450-T) = 150*2.6(T  - 50)

4500 - 10T = 390T<em> - 19500</em>

4500 + 19500 = 390T + 10T

24,000 = 400T

T = 24000/400

= 60⁰C

ΔSmetal = m1*c1In[t + 273]/[T1+273]

= 20*0.5 In (60+273)/450+273

= 10 ln(333/723)

= 10 * -0.7752

= -7.752

ΔS/oil =

m2*s2(60 + 273)/50 + 273)

= 150*2.6ln(333/323)

= 390 * 0.03048

= 11.88j/KgK

Δ<em>total = -7.7+11.8</em>

<em>= </em><em>4.18J/KgK</em>

this is the enthropy change

5 0
3 years ago
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