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garri49 [273]
3 years ago
14

A receptacle, plug, or any other electrical device whose design limits the ability of an electrician to come in contact with any

of its electrical connections is identified as _______ safe.
a. feel

b. resessed

c. touch

d. removal
Engineering
1 answer:
o-na [289]3 years ago
4 0

Answer:

Recessed  safe

Explanation:

A recessed, receptacle or plug is one in which the metal interfacing parts have been retracted back in such a way that common contact between a person handling the receptacle or outlet will not ordinarily lead to exposure to electric shock hazard and as such improving the safety of use of such plug, receptacle or electrical device either at home or in industrial setting

A recessed outlet is also more ideal than standard outlets where there are hazards such as potential contact with water.

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The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm
oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s

The exit temperature is:

T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25°C

b) The mean fluid velocity is:

V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s

The hydraulic diameter is:

D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m

The Reynold´s number is:

Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36

Assuming fully developed turbulent flow, the Nusselt number is:

Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69

h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C

The highest component surface temperature is:

T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22°C

6 0
3 years ago
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What
earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity (v), measured in meters per second, is determined by the following expression:

v = \frac{4\cdot \dot V}{\pi \cdot D^{2}} (1)

Where:

\dot V - Flow rate, measured in cubic meters per second.

D - Diameter, measured in meters.

If we know that \dot V = 0.01\,\frac{m^{3}}{s} and D = 0.05\,m, then the flow velocity is:

v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}

v \approx 5.093\,\frac{m}{s}

The density and dinamic viscosity of the glycerin at 20 ºC are \rho = 1260\,\frac{kg}{m^{3}} and \mu = 1.5\,\frac{kg}{m\cdot s}, then the Reynolds number (Re), dimensionless, which is used to define the flow regime of the fluid, is used:

Re = \frac{\rho\cdot v \cdot D}{\mu} (2)

If we know that \rho = 1260\,\frac{kg}{m^{3}}, \mu = 1.519\,\frac{kg}{m\cdot s}, v \approx 5.093\,\frac{m}{s} and D = 0.05\,m, then the Reynolds number is:

Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }

Re = 211.230

A pipeline is in turbulent flow when Re > 4000, otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor (f), dimensionless, is determined by the following expression:

f = \frac{64}{Re}

If we get that  Re = 211.230, then the friction factor is:

f = \frac{64}{211.230}

f = 0.303

The friction factor is 0.303.

4 0
2 years ago
Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He
Lyrx [107]
<h3>Answer:</h3>

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He works with the <u>Workers</u> , which consists of members who are building code officials and building safety professionals.

8 0
3 years ago
In a simple ideal Rankine cycle, water is used as the working fluid. The cycle operates with pressures of 2000 psi in the boiler
weqwewe [10]

Answer:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible,  therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is  not a given, so it cannot be calculated.

3 0
3 years ago
Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per
kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

                                                             $= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$. Then,

Power $ \propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

            $P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

           $P_2 = \text{const.} \times 1000 \ W$

5 0
3 years ago
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