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Alona [7]
3 years ago
11

Lockheed Martin Skunk Works designs and produces aircraft for defense using rapid prototyping tools

Engineering
1 answer:
Leni [432]3 years ago
4 0
Answer true


Explanation
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You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen
VashaNatasha [74]

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Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

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How is TEL (total equivalent length) measured and calculated? .​
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Measure the longest circuit and add 50% for fittings and terminal units.
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3 years ago
Biblical studies of john​
nevsk [136]

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<h2>the answer of sols brother is correct</h2><h3>hope it helps you have a good day</h3><h2 />

5 0
3 years ago
2.) A fluid moves in a steady manner between two sections in a flow
Talja [164]

Answer:

250\ \text{lbm/min}

625\ \text{ft/min}

Explanation:

A_1 = Area of section 1 = 10\ \text{ft}^2

V_1 = Velocity of water at section 1 = 100 ft/min

v_1 = Specific volume at section 1 = 4\ \text{ft}^3/\text{lbm}

\rho = Density of fluid = 0.2\ \text{lb/ft}^3

A_2 = Area of section 2 = 2\ \text{ft}^2

Mass flow rate is given by

m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}

The mass flow rate through the pipe is 250\ \text{lbm/min}

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}

The speed at section 2 is 625\ \text{ft/min}.

3 0
3 years ago
What is the definition of a duty cycle?
ira [324]

Answer:

D=\frac{PW}{T}*100

Explanation:

In electrical terms, is the ratio of time in which a load or circuit is ON compared to the time in which the load or circuit is OFF.

The duty cycle or power cycle, is expressed as a percentage of the activation time. For example, a 70% duty cycle is a signal that 70% of the time is activated and the other 30% disabled. Its equation can be expressed as:

D=\frac{PW}{T}*100

Where:

D=Duty\hspace{3}Cycle

PW=Pulse\hspace{3}Active\hspace{3}Time

T=Period\hspace{3}of\hspace{3}the\hspace{3}Signal

Here is a picture that will help you understand these concepts.

5 0
3 years ago
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