All categories

2 years ago

While discussing the testing of a Hall-effect switch:

Engineering

1 answer:

sergejj [24]2 years ago

6 0

Answer:

your answer is technician A

You might be interested in

Which of the following would be addressed by an employer completing an EAP template?

zloy xaker [14]

I can help you, but what are the options that were given to you?

4 0

2 years ago

What is the angular velocity (in rad/s) of a body rotating at N r.p.m.?

Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

$N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N$

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

3 0

3 years ago

7. Sockets internal designs come in what sizes?

MAXImum [283]

Answer:

These drive fittings come in four common sizes: 1⁄4 inch, 3⁄8 inch, 1⁄2 inch, and 3⁄4 inch (referred to as "drives", as in "3⁄8 drive").

3 0

2 years ago

An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is

baherus [9]

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft + 2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

4 0

3 years ago

A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.

AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass $m_{in}+m_{out}=\bigtriangleup m_{system}$

so, $m_e=m_1-m_2$

$Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1$

#Also, given the properties of water as;

$(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg$

#We assume constant properties for the steam at average temperatures: $h_e=\approx(h_1+h_2)/2$

#Replace known values in the equation above; $h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg$

#Using the mass and energy balance relations;

$m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg$

#We have $Q_i_n+m_eh_e=m_2u_2-m_1u_1$ : we replace the known values in the equation as;

$Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ$

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0

3 years ago