All categories

3 years ago

What was the Strategic Air Command?

2 answers:

6 0

I'm not sure UBS hand UBS d es. used ins d UBS d jbs is. DHS. u s dibs us dibs de DJ d DJ is end dijjee job DJ dje had d USB e ujjde in dr. hvdn. jdjhe

luda_lava [24]3 years ago

4 0

The Strategic Air Command was both a US Department of Defense Specified Command and a United States Air Force Major Command.

You might be interested in

How has dissection used in engineering?

NeTakaya

Answer:

Product dissection has been widely deployed in engineering education as a means to aid in student's understanding of functional product elements, development of new concept ideas, and their preparation for industry.

Explanation:

I hope this helps :) have a wonderful day!

6 0

3 years ago

Read 2 more answers

In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26

scZoUnD [109]

Answer:

The strength coefficient is $625$ and the strain-hardening exponent is $0.435$

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find $(K)$ and the $(n)$ .

$\sigma =K\epsilon^n$

We will plug the values in the formula.

$250=K\times (0.12)^n\\350=K\times (0.26)^n$

We will solve these equation.

$K=\frac{250}{(0.12)^n}$ plug this value in $350=K\times (0.26)^n$

$350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n$

Taking a natural log both sides we get.

$ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435$

Now, we will find value of $K$

$K=\frac{250}{(0.12)^n}$

$K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625$

So, the strength coefficient is $625$ and the strain-hardening exponent is $0.435$ .

5 0

3 years ago

Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per

Cloud [144]

Answer:

a) $Re_{D} = 111896.745$ , b) $Re_{D} = 1.119\times 10^{-7}$

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

$Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}$

Let assume that density and dynamic viscosity at 25 °C are $997\,\frac{kg}{m^{3}}$ $0.891\times 10^{-3}\,\frac{kg}{m\cdot s}$ , respectively. Then:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 111896.745$

b) The result is:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 1.119\times 10^{-7}$

6 0

4 years ago

.provide feedback to the controller about the robot's environment, a limited sense of sight and sound

anyanavicka [17]

Answer:

go to settings and u may figure out

6 0

4 years ago

Read 2 more answers

A beam spans 40 feet and carries a uniformly distributed dead load equal to 2.2 klf (not including beam self-weight) and a live

salantis [7]

Answer:

A beam is a structural member that is subjected primarily to transverse loads ... stress equal to σy, then the section Moment - Curvature (M-φ) response for .... Example 2.2 Design a simply supported beam subjected to uniformly distributed dead ... (live load). • Step I. Calculate the factored design loads (without self-weight).

Explanation:

6 0

3 years ago