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3 years ago

Is a street the same as a avenue

Engineering

2 answers:

-BARSIC- [3]3 years ago

5 0

they're essentially the same thing so i'd say yes

Whitepunk [10]3 years ago

5 0

Answer:

there kinda the same

Explanation:

You might be interested in

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

2 years ago

What does this middle button on the middle of my Chevrolet equinox steering wheel mean?

almond37 [142]

Answer:

Depending on how new your vehicle is, to me it looks like some sort of turn on button for forward collision safety feature, but i'm not an expert with this particular vehicle. If you want a better answer, I strongly suggest looking at the owner's manual under which should be located in the front dash compartment on the passengers side. Once you have the manual, look in the appendix until you find controls (or something similar) then go to that page and read about your vehicles control buttons. Your answer should be in the manual.

Something that all automobile owners should do right after purchasing a new vehicle, is reading the owners manual. As boring as it may seem, reading the owner's manual will help you get used to your new car quicker and give you instructions on how to take care of your car so that it lasts.

Have a great day, and I wish you safe traveling for now and forever! :)

7 0

3 years ago

Read 2 more answers

Por favor. alguien me comunique con fatimalisethmateual

lions [1.4K]

Wait why do you want me to

7 0

2 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

3 years ago

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the

muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, $\L_e$ = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

$P =\frac{\pi^2EI}{L_e^2}$

where, I is the moment of inertia

on substituting the respective values, we get

$10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}$

I = 13.13 in⁴

also for circular cross-section

I = $\frac{\pi}{64}\times d^4$

thus,

13.13 = $\frac{\pi}{64}\times d^4$

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0

3 years ago