All categories

3 years ago

B is to 10 as 7 is to 2 setting porption

Engineering

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

im so sorry about that girl

Explanation:

You might be interested in

A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 1

Naily [24]

Answer:

a) attached below

b) 0.0337

c) 2730.206 Ib

d) 2320.338 ft/min

Explanation:

<u>a) Plot of the drag polar for this aircraft </u>

first we will calculate :

Wing area (s) = Wing span (b) * Average chord length(c)

= 53.3 * 6 = 319.8 ft^2

Aspect ratio = b^2 / s = 8.883

K = 1 / $\pi$ eAR = 1 /

Drag polar ( Cd ) = 0.02 + 0.044 C^2L

attached below is a plot of the drag polar

<u />

Attached below is the detailed solution of the remaining part of the question

6 0

3 years ago

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

Which of the following is true about how the universe is expanding?

Law Incorporation [45]

B. The space between the galaxy is getting better

4 0

3 years ago

Read 2 more answers

A voltage regulator is to provide a constant DC voltage Vl=10V to a load Rl from a nominal Vcc=15V supply voltage. The load can

Luden [163]

Answer:

Beta values can be from the equation=change in Vcc/nominal Vcc

Beta=16-3/15=3/15=1/5=0.20

Maximum power=I^2*R=40 W

7 0

3 years ago

Roku internet service providet

podryga [215]

Answer:

What?? I do not understand?

6 0

3 years ago