2 years ago

What are four engineering degrees that can help lead you to becoming an aerospace engineer

Engineering

1 answer:

mamaluj [8]2 years ago

7 0

Answer:

Mechanical Engineering, Aerospace Engineering, Electrical Engineering, Aeronautical Engineering

Explanation:

I'm a Mechanical Engineer, and I considered Aerospace.

You might be interested in

Which of the following are true about the American Wire Gauge?

harina [27]

Answer:

A. smallest wire is No. 12

7 0

2 years ago

You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,

nadezda [96]

Answer:

Explanation:

cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

= 1.075 x 10⁻³

Young's modulus = stress / strain

= 83.32 x 10⁶ / 1.075 x 10⁻³

= 77.5 x 10⁹ N/m²

5 0

3 years ago

The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol

pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = $\frac{\pi}{4}D^2L$

V = $\frac{\pi}{4}\frac{D^3}{1.1}$

on substituting the values, we have

245 = $\frac{\pi}{4}\frac{D^3}{1.1}$

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

L= 6.36 cm

Now,

the compression ratio is given as:

$\textup{compression ratio}=\frac{V+v}{v}$

on substituting the values, we get

$\textup{compression ratio}=\frac{245+27.2}{27.2}$

Compression ratio = 10.007

4 0

3 years ago

Everyone why are you reporting my answers i didnt do anything to you

masya89 [10]

Answer:

IDK

Explanation:

same thing is happening to me

5 0

3 years ago

Read 2 more answers

List, in ascending order, the cutoff frequencies for the first ten modes of a rectangular waveguide, normalized to the cutoff fr

Alisiya [41]

Answer:

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

Explanation:

Assuming a = 2b

Attached below is the required steps to the solution

The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order is :

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

5 0

3 years ago