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elena-s [515]
3 years ago
9

True or false? Engineering degree programs and engineering technology degree programs have a different requirements

Engineering
1 answer:
sergey [27]3 years ago
4 0
That’s true brother
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(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced
Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

See attached image file

5 0
3 years ago
Their game off badminton is always on Tuesday
igomit [66]

Ok? is this a question or what?

5 0
2 years ago
A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str
just olya [345]

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is \frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}

The shear strength=0.58\sigma_y and in this case \sigma_y=36 000 psi

Shear strength=\frac {Load}{Shear area} hence making load the subject then

Load=Shear area X Shear strength

Load=\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi

3 0
3 years ago
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an
ser-zykov [4K]

Answer:

volumetric flow rate = 0.0251 m^3/s

Velocity in pipe section 1 = 6.513m/s

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = 0.0251 m^3/s

volumetric flow rate = 0.0251 m^3/s

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s

<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s

7 0
3 years ago
Are there engineering students here?​
leva [86]
Uh, I’d assume so because Brainly has a whole section of questions for them.
7 0
3 years ago
Read 2 more answers
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