Chemical engineering is one of the simpler fields in engineering.<br> True<br> False

Which of the following results from the fact that skip signals refracted from the ionosphere are elliptically polarized?

zhenek [66]

Answer:

B

Explanation:

The fact that skip signals refracted from the ionosphere are elliptically polarized is a result of either vertically or horizontally polarized antennas may be used for transmission or reception.

8 0

3 years ago

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the

Olegator [25]

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = $\frac{\pi}{4}(D^2-d^2)$

or

Area of hollow pipe = $\frac{\pi}{4}(D^2-100^2)$

thus,

400 × 10³ N = 120 × $\frac{\pi}{4}(D^2-100^2)$

or

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

or

D = 119.35 mm

7 0

3 years ago

Read 2 more answers

4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape

snow_lady [41]

Answer:

2102.1 m

Explanation:

Temperature at the equator = 0⁰

Radius of the earth = 6.37x10⁶

Required:

We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.

Final temperature = ∆T = 303-273 = 30

S = 11x10^-6

The clearance R = Ro*S*∆T

=6.37x10⁶x 11x10^-6x30

= 2102.1m

Or 2.102 kilometers

Thank you

4 0

3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$