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3 years ago

6

3. Determine the most unfavorable arrangement of the crane loads and

1 answer:

Gnoma [55]3 years ago

3 0

Answer:

There are 350cm in 3.5m.

Explanation:

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Which of the following is the next “up-and-coming” digital publishing industry?

storchak [24]

Answer:

3D printing

Explanation:

The developments in technology, infrastructure, systems and technology mean that 3D printing will soon become a production technology. The market penetration of 3D printing would inevitably rise over time, with certain categories almost completely transitioning to 3D printing, such as digital publishing.

6 0

3 years ago

What is the built-in pollution control system in an incinerator called

Kobotan [32]

Explanation:

hbyndbnn☝️

7 0

3 years ago

A nanometer connected to a pipe indicates a negative gauge pressure of 70mm of mercury .what is the pressure in the pipe in N/m^

prohojiy [21]

Based on the calculations, the absolute pressure in the pipe is equal to 90,670.4‬ N/m².

<h3>How to calculate the absolute pressure?</h3>

Mathematically, absolute pressure can be calculated by using this formula:

$P_{abs} = P_{atm} + P_{guage}$

<u>Scientific data:</u>

Atmospheric pressure (Patm) = 1 bar = 1 × 10⁵ N/m²

Head (h) = -70mmhg = -0.07 m hg

Acceleration due to gravity = 9.8 m/s²

Density of mercury = 13,600 kg/m³.

Next, we would determine the gauge pressure:

$P_{guage} = \rho gh\\\\P_{guage} = -13600 \times 9.8 \times 0.07\\\\P_{guage} = -9,329.6\;N/m^2$

Now, we can calculate the absolute pressure:

$P_{abs} = P_{atm} + P_{guage}\\\\P_{abs} = 1 \times 10^5 - 9329.6$

Absolute pressure = ‭90,670.4‬ N/m².

Read more on absolute pressure here: brainly.com/question/10013312

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6 0

2 years ago

Who invented a control unit for an artificial heart?<br> ements<br> ante

disa [49]

Otis Boykin is the answer.

6 0

3 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

4 years ago