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3 years ago

8. True or False? Private companies are involved in the New Space Race.

Engineering

1 answer:

yan [13]3 years ago

5 0

The answer is true. Elon musk has a company called space x

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The two types of outlets that are found in an electrical system are:______

Dmitry [639]

Answer:

a. lighting and receptacle outlets

Explanation:

The two types of outlets that are found in an electrical system are

a. lighting and receptacle outlets

Outlets allow electrical equipment to connect to the electrical grid. The electrical grid provides alternating current to the outlet.

3 0

3 years ago

The type of current that flows from the electrode across the arc to the work is called what?

Scrat [10]

Answer:

Direct current.

Explanation:

5 0

3 years ago

An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is

baherus [9]

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft + 2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

4 0

3 years ago

Impact strips may be designed into a bumper cover.<br> True<br> False

d1i1m1o1n [39]

Answer:

true I think

Explanation:

true I think

5 0

3 years ago

Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E

White raven [17]

Answer:

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

Explanation:

To solve this problem we use the expression for the temperature film

$T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5$

Then, we have to compute the Reynolds number

$Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}$

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

$Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77$

but we also now that

$Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\$

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0

4 years ago