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2 years ago

PLEASE FIX THIS LUA SCRIPT

Engineering

2 answers:

yKpoI14uk [10]2 years ago

6 0

Answer: not sure

Explanation: brainly please

zavuch27 [327]2 years ago

5 0

Answer:

When declaring variables, you said “workspace.MeanEnemy”. What you should have done was “game.Workspace.MeanEnemy”. Same for the variable right under it. Other than that, seems to be all good!

Hope This Helped!

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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

A strip of chicken skin was excised for mechanical testing in tension. The initial dimensions of the rectangular specimen were 3

Stells [14]

Answer:

Table and chart are attached below

Explanation:

4 0

4 years ago

Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin

hammer [34]

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

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4 years ago

Some designers suggest that speech recognition should be used in a telephone menu system. This would allow users to interact wit

Ghella [55]

The designers suggest that speech recognition allows for spoken interaction or conversation and spoken prompts or commands.

Explanation:

The speech recognition is used when the user has physical impairments and hands are busy, eyes are occupied and the user cannot read.

We can save time with the usage of speech recognition system. The users are knowledgeable based on the actions available.

The problems faced in speech recognition system are in the noisy environment it has bad microphones and the commands should be learned and remembered.

The other obstacle is error correction is time consuming. the speech production has slow space to speech output provides privacy in public spaces. The disadvantage is it contains large amount of information.

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4 years ago

If the gear ratio is 3:5 does it increase torque or speed?

Bas_tet [7]

Answer:

yes it does

Explanation:

3 0

3 years ago

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