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3 years ago

When you arrive at an intersection with a stop sign in your direction, if there is no marked stop

Engineering

2 answers:

kakasveta [241]3 years ago

6 0

Answer:

The answer is B. Wait for the pedestrians in the crosswalk to cross, then stop on the pedestrian crosswalk.

Illusion [34]3 years ago

3 0

Answer:

C: Stop before entering the pedestrian crosswalk.

Explanation:

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For laminar flow over a flat plate, the local heat transfer coefficient hx is known to vary as x−1/2, where x is the distance fr

Reika [66]

Answer:

Explanation:

So for solving this problem we need the local heat transfer coefficient at distance x,

$h_x=cx^{-1/2}$

We integrate between 0 to x for obtain the value of the coefficient, so $\bar{h}_x =\frac{1}{x} \int\limit^x_0 h_x dx\\\bar{h}_x = \frac{c}{x} \int\limit^x_0 \frac{1}{\sqrt{x}}dx\\\bar{h}_x = \frac{c}{c} (2x^{1/2})\\\bar{h}_x = 2cx^{-1/2}$

Substituing

$\bar{h}_x=2h_x\\\frac{\bar{h}_x}{h_X}=2$

The ratio of the average convection heat transfer coefficient over the entire length is 2

6 0

4 years ago

Plane wall of material A with internal heat generation is insulated on one side and bounded by a second wall of material B, whic

viktelen [127]

Sorry❤

Have a nice day ✨

8 0

3 years ago

What caused the rate of erosion to change around 1905 and 1906?

weqwewe [10]

Answer:

Building the Canadian hydroelectric plant reduced the rate of erosion after 1906. This change in the rate of erosion is represented by the change in the slope on the graph. Hydroelectric dams convert kinetic energy from moving water to electrical energy.

4 0

3 years ago

Read 2 more answers

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:

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